Chapter 11 Electrostatics Solved Numericals (30 Questions with Solutions)

Chapter 11 Electrostatics Solved Numericals – Class 12 Physics

Welcome to the complete collection of Chapter 11 Electrostatics solved numericals. This comprehensive resource includes 30 carefully selected numerical problems arranged from basic concepts to advanced competitive-level questions, making it ideal for students preparing for Board Examinations, MDCAT, ECAT, NUST, PIEAS, GIKI, UET, FAST, and other engineering or medical entrance tests.

Each numerical is solved using a clear, step-by-step method that includes the given data, required quantity, relevant formula, detailed calculations, final answer, concept explanation, and exam tips. This structured approach helps students strengthen their understanding of electrostatics while improving problem-solving skills and examination confidence.

Whether you are revising before your annual examinations or preparing for competitive entry tests, these solved numericals provide an excellent practice resource to master Coulomb's Law, Electric Field, Electric Potential, Electric Flux, Gauss's Law, Capacitance, Capacitors, Electric Energy, and Electrostatic Applications.

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Solved Numerical 11.1

Comparison of Electrical and Gravitational Forces

Difficulty Level: 🟢 Easy

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Problem

Compare the magnitudes of the electrical force and the gravitational force exerted on an object of mass 20 g and charge 20 μC by an identical object placed 10 cm away. Take

$$G=6.67\times10^{-11}\;Nm^2kg^{-2}$$

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Given

$$m=20g=0.02kg$$
$$q=20\mu C=20\times10^{-6}C$$
$$r=10cm=0.1m$$
$$k=9\times10^9Nm^2C^{-2}$$

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Required

Compare

• Electrical Force
• Gravitational Force

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Formula

Electrical Force

$$F_e=k\frac{q^2}{r^2}$$

Gravitational Force

$$F_g=G\frac{m^2}{r^2}$$

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Solution

Step 1: Electrical Force

$$F_e= 9\times10^9 \left( \frac{(20\times10^{-6})^2}{(0.1)^2} \right)$$
$$= 9\times10^9 \left( \frac{4\times10^{-10}}{0.01} \right)$$
$$= 360N$$

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Step 2: Gravitational Force

$$F_g= 6.67\times10^{-11} \frac{(0.02)^2}{(0.1)^2}$$
$$= 6.67\times10^{-11} \left( \frac{4\times10^{-4}}{0.01} \right)$$
$$= 2.67\times10^{-12}N$$

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Step 3: Comparison

$$\frac{F_e}{F_g} = \frac{360}{2.67\times10^{-12}}$$
$$= 1.35\times10^{14}$$

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Final Answer

Electrical Force

$$\boxed{F_e=360N}$$

Gravitational Force

$$\boxed{F_g=2.67\times10^{-12}N}$$

Comparison

$$\boxed{ F_e\approx1.35\times10^{14}F_g }$$

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Concept Behind the Numerical

The electrical force between charged particles is enormously greater than the gravitational force between the same particles.

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Board Exam Tip

Always convert:

• grams → kilograms
• centimetres → metres
• microcoulombs → coulombs

before substitution.

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Solved Numerical 11.2

Electric Field on the z-axis

Difficulty Level: 🟢 Easy

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Problem

A point charge q=−8×10⁻⁸ C is placed at the origin. Calculate the electric field at a point 2 m from the origin on the z-axis.

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Given

$$q=-8\times10^{-8}C$$
$$r=2m$$
$$k=9\times10^9Nm^2C^{-2}$$

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Required

Electric Field Intensity

---

Formula

$$E=k\frac{q}{r^2}$$

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Solution

$$E= 9\times10^9 \left( \frac{-8\times10^{-8}}{2^2} \right)$$
$$= 9\times10^9 \times2\times10^{-8}$$
$$= 180N/C$$

Since charge is negative,

the field points towards the origin.

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Final Answer

$$\boxed{ E=180N/C }$$

Direction:

$$\boxed{ \text{Towards the negative charge} }$$

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Concept Behind the Numerical

Electric field due to a negative charge always points towards the charge.

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Board Exam Tip

Magnitude is always positive.

Direction should be mentioned separately.

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Solved Numerical 11.3

Electric Field at a Given Position

Difficulty Level: 🟢 Easy

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Problem

Determine the electric field at

$$r=(4\widehat{i}+3\widehat{j})m$$

caused by a point charge

$$q=5\times {10}^{-6}C$$

located at the origin.

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Given

$$q=5\times10^{-6}C$$
$$r=(4\hat i+3\hat j)m$$

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Required

Electric Field

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Formula

Distance

$$r=\sqrt{x^2+y^2}$$

Electric Field

$$E=k\frac{q}{r^2}$$

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Solution

Distance

$$r=\sqrt{4^2+3^2}$$
$$=5m$$

Electric Field

$$E= 9\times10^9 \frac{5\times10^{-6}}{25}$$
$$= 1800N/C$$

Direction

$$\hat r= \frac{4\hat i+3\hat j}{5}$$

Therefore,

$$\vec E = 1800 \left( \frac45\hat i+\frac35\hat j \right)$$
$$= 1440\hat i+1080\hat j$$

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Final Answer

Magnitude

$$\boxed{1800N/C}$$

Vector Form

$$\boxed{ \vec E = 1440\hat i + 1080\hat j }$$

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Concept Behind the Numerical

Electric field is a vector quantity.

Both magnitude and direction are important.

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Solved Numerical 11.4

Force on a Test Charge

Difficulty Level: 🟢 Easy

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Problem

A charge 3 μC is placed in an electric field of 250 N/C. Find the force acting on it.

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Given

$$q=3\times10^{-6}C$$
$$E=250N/C$$

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Required

Force

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Formula

$$F=qE$$

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Solution

$$F= 3\times10^{-6} \times250$$
$$= 7.5\times10^{-4}N$$

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Final Answer

$$\boxed{ F=7.5\times10^{-4}N }$$

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Concept Behind the Numerical

A charge placed in an electric field experiences a force proportional to both its charge and the field strength.

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Solved Numerical 11.5

Electric Field at Different Distances

Difficulty Level: 🟢 Easy

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Problem

A charge 6μC produces an electric field. Calculate the electric field at a distance of 0.5 m.

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Given

$$q=6\times10^{-6}C$$
$$r=0.5m$$

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Required

Electric Field

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Formula

$$E=k\frac{q}{r^2}$$

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Solution

$$E= 9\times10^9 \frac{6\times10^{-6}}{(0.5)^2}$$
$$= 9\times10^9 \frac{6\times10^{-6}}{0.25}$$
$$= 2.16\times10^5N/C$$

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Final Answer

$$\boxed{ E=2.16\times10^5N/C }$$

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Concept Behind the Numerical

Electric field decreases according to the inverse square law. If the distance doubles, the field becomes one-fourth of its original value.

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Board Exam Tip

Remember the common relation:

• Double the distance → Electric field becomes ¼
• Triple the distance → Electric field becomes 1⁄9

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Solved Numerical 11.6

Electric Dipole Moment

Difficulty Level: 🟢 Easy

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Problem

Two equal and opposite charges of magnitude 5 μC are separated by a distance of 8 cm. Calculate the electric dipole moment.

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Given

$$q=5\times10^{-6}C$$
$$d=8cm=0.08m$$

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Required

Electric Dipole Moment

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Formula

$$p=qd$$

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Solution

Substituting the given values,

$$p=(5\times10^{-6})(0.08)$$
$$=4\times10^{-7}Cm$$

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Final Answer

$$\boxed{p=4\times10^{-7}\;Cm}$$

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Concept Behind the Numerical

The electric dipole moment depends upon:

• magnitude of charge
• separation between charges

Greater separation produces a larger dipole moment.

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Board Exam Tip

Always convert cm into metres before using the formula.

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Solved Numerical 11.7

Electric Flux Through a Surface

Difficulty Level: 🟢 Easy

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Problem

A uniform electric field of 500 N C⁻¹ makes an angle of 60° with the normal to a flat surface of area 0.20 m². Calculate the electric flux through the surface.

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Given

$$E=500N/C$$
$$A=0.20m^2$$
$$\theta=60^\circ$$

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Required

Electric Flux

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Formula

$$\Phi=EA\cos\theta$$

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Solution

$$\mathrm{\Phi }=500\times 0.20\times \cos {60}^{\circ }$$

Since

$$\cos60^\circ=0.5$$
$$\Phi = 500 \times 0.20 \times 0.5$$
$$=50Nm^2/C$$

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Final Answer

$$\boxed{\Phi=50Nm^2/C}$$

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Concept Behind the Numerical

Electric flux depends upon

• electric field
• area
• orientation of the surface.

Maximum flux occurs when the field is perpendicular to the surface.

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Board Exam Tip

Do not use the angle with the surface itself.

Always use the angle with the normal.

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Solved Numerical 11.8

Suspended Charged Particle Between Two Plates

Difficulty Level: 🟡 Moderate

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Problem

Find the electric field strength required to suspend a particle of mass

$$1\times {10}^{-6}kg$$

having charge

$$1\mu C$$

between two horizontal plates.

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Given

$$m=1\times10^{-6}kg$$
$$q=1\times10^{-6}C$$
$$g=9.8m/s^2$$

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Required

Electric Field Strength

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Formula

For equilibrium,

$$qE=mg$$

Hence,

$$E=\frac{mg}{q}$$

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Solution

$$E = \frac{(1\times10^{-6})(9.8)} {1\times10^{-6}}$$
$$=9.8N/C$$

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Final Answer

$$\boxed{E=9.8N/C}$$

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Concept Behind the Numerical

When a charged particle remains suspended,

Electric Force = Weight

$$qE=mg$$

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Board Exam Tip

Notice that the plate separation is not required in this calculation.

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Solved Numerical 11.9

Energy Gained by Electrons

Difficulty Level: 🟡 Moderate

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Problem

A particle carrying a charge equal to 20 electrons moves through a potential difference of 100 V. Calculate the energy gained by the particle in

(a) electron volts

(b) joules.

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Given

Number of electrons

$$n=20$$

Charge on one electron

$$e=1.6\times10^{-19}C$$

Potential Difference

$$V=100V$$

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Required

Energy gained

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Formula

Total charge

$$Q=ne$$

Energy

$$W=QV$$

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Solution

Charge carried

$$Q = 20 \times 1.6\times10^{-19}$$
$$=3.2\times {10}^{-18}C$$

Energy

$$W = 3.2\times10^{-18} \times100$$
$$= 3.2\times10^{-16}J$$

Since one electron crossing 100 V gains 100 eV,

Twenty electrons gain

$$20\times100 = 2000eV$$

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Final Answer

In electron volts

$$\boxed{2000eV}$$

In joules

$$\boxed{3.2\times10^{-16}J}$$

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Concept Behind the Numerical

One electron gains 1 eV when accelerated through 1 volt.

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Board Exam Tip

For energy in eV,

Multiply

Number of electrons × Potential Difference.

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Solved Numerical 11.10

Electric Potential Due to a Point Charge

Difficulty Level: 🟡 Moderate

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Problem

Using infinity as the reference point, determine the electric potential at a point 1.2 m away from a point charge 4×10⁻⁸ C for:

(a) positive charge

(b) negative charge.

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Given

$$q=\pm4\times10^{-8}C$$
$$r=1.2m$$
$$k=9\times10^9Nm^2C^{-2}$$

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Required

Electric Potential

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Formula

$$V=k\frac{q}{r}$$

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Solution

(a) Positive Charge

$$V = 9\times10^9 \left( \frac{4\times10^{-8}}{1.2} \right)$$
$$= 300V$$

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(b) Negative Charge

The magnitude remains the same,

but the sign becomes negative.

$$V=-300V$$

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Final Answer

For positive charge

$$\boxed{V=+300V}$$

For negative charge

$$\boxed{V=-300V}$$

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Concept Behind the Numerical

Electric potential is a scalar quantity.

Its sign depends only on the nature of the source charge.

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Board Exam Tip

Many students forget to include the negative sign for a negative source charge. Always check whether the source charge is positive or negative before writing the final answer.

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Solved Numerical 11.11

Millikan's Oil Drop Experiment

Difficulty Level: 🟡 Moderate

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Problem

In Millikan's oil drop experiment, oil droplets are introduced between two horizontal plates 500 mm apart. The potential difference between the plates is adjusted to 780 V, so that the oil droplet remains suspended. When the electric field is removed, the droplet falls 1.50 mm in 11.2 s.

Given:

• Density of oil = 900 kg m⁻³
• Viscosity of air = 1.8 × 10⁻⁵ N s m⁻²
• $g=9.8\,m\,s^{-2}$
  

Calculate:

(a) Mass of the droplet

(b) Charge on the droplet.

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Given

Plate separation

$$d=500mm=0.5m$$

Potential difference

$$V=780V$$

Density

$$\rho=900kg/m^3$$

Viscosity

$$\eta=1.8\times10^{-5}Ns/m^2$$

Distance fallen

$$s=1.50\times {10}^{-3}m$$

Time

$$t=11.2s$$

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Required

(a) Mass

(b) Charge

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Formula

Terminal velocity

$$v=\frac{s}{t}$$

Radius (Stokes' Law)

$$r=\sqrt{\frac{9\eta v}{2\rho g}}$$

Mass

$$m=\frac43\pi r^3\rho$$

Electric field

$$E=\frac{V}{d}$$

Charge

$$q=\frac{mg}{E}$$

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Solution

Terminal velocity

$$v=\frac{1.50\times10^{-3}}{11.2}$$
$$=1.34\times10^{-4}m/s$$

Radius

$$r=3.49\times10^{-6}m$$

Mass

$$m=1.61\times10^{-13}kg$$

Electric field

$$E=\frac{780}{0.5} =1560N/C$$

Charge

$$q=\frac{mg}{E}$$
$$=\frac{(1.61\times10^{-13})(9.8)}{1560}$$
$$=1.01\times10^{-15}C$$

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Final Answer

Mass

$$\boxed{1.61\times10^{-13}kg}$$

Charge

$$\boxed{1.01\times10^{-15}C}$$

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Concept Behind the Numerical

Millikan's experiment proved that electric charge is quantized.

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Board Exam Tip

This numerical involves several steps. Always write each formula separately and keep SI units throughout.

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Solved Numerical 11.12

Proton Moving in a Uniform Electric Field

Difficulty Level: 🟡 Moderate

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Problem

A proton is placed in a uniform electric field of 5000 N C⁻¹ directed towards the right. It moves 10 cm from point A to point B.

Calculate:

(a) Potential Difference

(b) Work Done

(c) Change in Potential Energy

(d) Change in Kinetic Energy

(e) Final Velocity

Mass of proton

$$m=1.67\times10^{-27}kg$$

Charge of proton

$$q=1.6\times10^{-19}C$$

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Given

$$E=5000N/C$$
$$d=0.10m$$
$$q=1.6\times10^{-19}C$$

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Formula

Potential Difference

$$V=Ed$$

Work Done

$$W=qV$$

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Solution

Potential Difference

$$V=5000(0.10)$$
$$=500V$$

Work Done

$$W=(1.6\times10^{-19})(500)$$
$$=8\times10^{-17}J$$

Potential energy decreases

$$\Delta U=-8\times10^{-17}J$$

Kinetic energy increases

$$\mathrm{\Delta }K=8\times {10}^{-17}J$$

Velocity

$$K=\frac12mv^2$$
$$v=\sqrt{\frac{2K}{m}}$$
$$v=3.09\times10^5m/s$$

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Final Answer

(a)

$$\boxed{500V}$$

(b)

$$\boxed{8\times10^{-17}J}$$

(c)

$$\boxed{-8\times10^{-17}J}$$

(d)

$$\boxed{8\times10^{-17}J}$$

(e)

$$\boxed{3.09\times10^5m/s}$$

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Board Exam Tip

Remember

Positive charge moving along the electric field

➡ Potential Energy decreases

➡ Kinetic Energy increases

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Solved Numerical 11.13

Electric Potential Due to a Point Charge

Difficulty Level: 🟢 Easy

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Problem

Using infinity as the reference point, determine the electric potential at a point 1.2 m from a charge 4×10⁻⁸C for

(a) Positive charge

(b) Negative charge

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Solution

Formula

$$V=\frac{kq}{r}$$

Substitute

$$V= \frac{(9\times10^9)(4\times10^{-8})}{1.2}$$
$$=300V$$

Negative charge

$$V=-300V$$

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Final Answer

Positive charge

$$\boxed{+300V}$$

Negative charge

$$\boxed{-300V}$$

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Concept

Electric potential is positive around positive charges and negative around negative charges.

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Solved Numerical 11.14

Bohr's Hydrogen Atom

Difficulty Level: 🔴 Challenging

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Problem

In Bohr's atomic model of the hydrogen atom, the electron revolves around the nucleus in a circular orbit of radius 5.29×10⁻¹¹m with speed 2.18×10⁶ m/s

Calculate

(a) The electric potential produced by the proton at the position of the electron.

(b) The total energy of the hydrogen atom in electron volts (eV).

(c) The ionization energy of the hydrogen atom.

Take

$$k=9.0\times10^9\,Nm^2C^{-2}$$
$$e=1.6\times {10}^{-19}C$$

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Given

Radius of orbit

$$𝑟=5.29\times {10}^{-11}𝑚$$

Speed of electron

$$𝑣=2.18\times {10}^6𝑚\mathrm{/}𝑠$$

Charge of proton

$$𝑞=1.6\times {10}^{-19}𝐶$$

Coulomb's constant

$$𝑘=9.0\times {10}^9𝑁{𝑚}^2{𝐶}^{-2}$$

Required

Calculate:

• Electric potential
• Total energy
• Ionization energy

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Formulae

Electric Potential

$$𝑉=\frac{𝑘𝑞}{𝑟}$$

Potential Energy

$$𝑈=-\frac{𝑘{𝑞}^2}{𝑟}$$

Kinetic Energy

$$𝐾=\frac{1}{2}𝑚{𝑣}^2$$

Total Energy

$$𝐸=𝐾+𝑈$$

For the ground state of hydrogen,

$$𝐸=-13.6𝑒𝑉$$

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Solution

(a) Electric Potential

Using

$$𝑉=\frac{𝑘𝑞}{𝑟}$$

Substitute the given values:

$$𝑉=\frac{(9.0\times {10}^9)(1.6\times {10}^{-19})}{5.29\times {10}^{-11}}$$
$$𝑉=27.2𝑉$$

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(b) Total Energy

For the ground state of the hydrogen atom,

Potential Energy

$$𝑈=-27.2𝑒𝑉$$

Kinetic Energy

$$𝐾=+13.6𝑒𝑉$$

Therefore,

$$𝐸=𝐾+𝑈$$
$$𝐸=13.6-27.2$$
$$𝐸=-13.6𝑒𝑉$$

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(c) Ionization Energy

Ionization energy is the energy required to remove the electron completely from the ground state.

Therefore,

$$\boxed{{\displaystyle \text{Ionization Energy}=13.6𝑒𝑉}}$$

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Final Answer

(a) Electric Potential

$$\boxed{{\displaystyle 𝑉=27.2𝑉}}$$

(b) Total Energy

$$\boxed{{\displaystyle 𝐸=-13.6𝑒𝑉}}$$

(c) Ionization Energy

$$\boxed{{\displaystyle 13.6𝑒𝑉}}$$

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Concept Behind the Numerical

In the ground state of the hydrogen atom:

• The proton produces an electric potential of 27.2 V at the electron's orbit.
• The electron possesses 13.6 eV of kinetic energy.
• Its potential energy is −27.2 eV.
• The total energy is −13.6 eV, indicating that the electron is bound to the nucleus.
• An energy of 13.6 eV must be supplied to completely remove the electron from the atom.

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Board Examination Tips

✅ Memorize these standard values for the hydrogen atom (ground state):

Quantity	Value
Orbit Radius	$5.29\times {10}^{-11}𝑚$
Electron Speed	$2.18\times {10}^6𝑚\mathrm{/}𝑠$
Electric Potential	27.2 V
Potential Energy	−27.2 eV
Kinetic Energy	+13.6 eV
Total Energy	−13.6 eV
Ionization Energy	13.6 eV

These values are frequently asked directly in Board Examinations, MDCAT, ECAT, NUST, PIEAS, GIKI, UET, and FAST and are worth memorizing. They also make excellent one-mark conceptual questions.

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Solved Numerical 11.15

Electrons Transferred to a Capacitor

Difficulty Level: 🟡 Moderate

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Problem

A capacitor has capacitance 2.5×10⁻⁸ F. When connected to a source of 450 V, determine the number of electrons transferred.

Take

$$e=1.6\times10^{-19}C$$

---

Given

$$C=2.5\times10^{-8}F$$
$$V=450V$$

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Formula

Charge

$$Q=CV$$

Number of electrons

$$n=\frac{Q}{e}$$

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Solution

Charge

$$Q=(2.5\times10^{-8})(450)$$
$$=1.125\times10^{-5}C$$

Number of electrons

$$n = \frac{1.125\times10^{-5}} {1.6\times10^{-19}}$$
$$=7.03\times10^{13}$$

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Final Answer

Charge

$$\boxed{1.125\times10^{-5}C}$$

Electrons transferred

$$\boxed{7.03\times10^{13}}$$

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Concept Behind the Numerical

Charging a capacitor transfers electrons from one plate to the other, creating equal and opposite charges.

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Board Exam Tip

Always calculate the charge first using

$$Q=CV$$

and then determine the number of electrons using

$$n=\frac{Q}{e}$$

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Solved Numerical 11.16

Capacitance of a Parallel Plate Capacitor

Difficulty Level: 🟢 Easy

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Problem

A parallel plate capacitor has plate area 0.020 m² and plate separation 2 mm. Calculate its capacitance.

Take

$$\varepsilon_0=8.85\times10^{-12}F/m$$

---

Given

$$A=0.020m^2$$
$$d=2mm=2\times10^{-3}m$$
$$\varepsilon_0=8.85\times10^{-12}F/m$$

---

Required

Capacitance

---

Formula

$$C=\frac{\varepsilon_0A}{d}$$

---

Solution

Substitute the values:

$$C= \frac{(8.85\times10^{-12})(0.020)} {2\times10^{-3}}$$
$$C=8.85\times10^{-11}F$$

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Final Answer

$$\boxed{C=8.85\times10^{-11}F}$$

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Concept Behind the Numerical

Capacitance increases with larger plate area and decreases with greater plate separation.

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Board Exam Tip

Always convert mm into metres before substitution.

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Solved Numerical 11.17

Charge Stored on a Capacitor

Difficulty Level: 🟢 Easy

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Problem

A capacitor of capacitance 8 μF is connected across a 120 V battery. Find the charge stored on the capacitor.

---

Given

$$C=8\mu F =8\times10^{-6}F$$
$$V=120V$$

---

Required

Charge stored

---

Formula

$$Q=CV$$

---

Solution

$$Q = (8\times10^{-6})(120)$$
$$Q=9.6\times10^{-4}C$$

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Final Answer

$$\boxed{Q=9.6\times10^{-4}C}$$

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Concept Behind the Numerical

A capacitor stores charge directly proportional to both its capacitance and applied voltage.

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Board Exam Tip

Remember the simple relationship:

$$Q=CV$$

This is one of the most frequently used equations in capacitor numericals.

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Solved Numerical 11.18

Energy Stored in a Capacitor

Difficulty Level: 🟡 Moderate

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Problem

A capacitor of capacitance 12 μF is charged to a potential difference of 200 V. Calculate the energy stored in the capacitor.

---

Given

$$C=12\times10^{-6}F$$
$$V=200V$$

---

Required

Energy Stored

---

Formula

$$U=\frac12CV^2$$

---

Solution

$$U = \frac12 (12\times10^{-6}) (200)^2$$
$$= \frac12 (12\times10^{-6}) (40000)$$
$$U=0.24J$$

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Final Answer

$$\boxed{U=0.24J}$$

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Concept Behind the Numerical

A charged capacitor stores electrical energy in the electric field between its plates.

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Board Exam Tip

When voltage is given directly, the quickest formula is

$$U=\frac12CV^2$$

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Solved Numerical 11.19

Capacitors Connected in Series

Difficulty Level: 🟡 Moderate

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Problem

Three capacitors of capacitances 4 μF, 6 μF, and 12 μF are connected in series. Find the equivalent capacitance.

---

Given

$$C_1=4\mu F$$
$$C_2=6\mu F$$
$$C_3=12\mu F$$

---

Required

Equivalent Capacitance

---

Formula

$$\frac1C= \frac1{C_1} + \frac1{C_2} + \frac1{C_3}$$

---

Solution

$$\frac1C = \frac14 + \frac16 + \frac1{12}$$

Taking LCM = 12

$$\frac{1}{C}=\frac{3}{12}+\frac{2}{12}+\frac{1}{12}=\frac{6}{12}=\frac{1}{2}$$

Therefore,

$$C=2\mu F$$

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Final Answer

$$\boxed{C=2\mu F}$$

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Concept Behind the Numerical

In series:

• Charge remains the same.
• Equivalent capacitance is always less than the smallest capacitor.

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Board Exam Tip

This question is asked very frequently in board examinations.

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Solved Numerical 11.20

Capacitors Connected in Parallel

Difficulty Level: 🟢 Easy

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Problem

Three capacitors having capacitances 5 μF, 8 μF, and 10 μF are connected in parallel. Find the equivalent capacitance.

---

Given

$$C_1=5\mu F$$
$$C_2=8\mu F$$
$$C_3=10\mu F$$

---

Required

Equivalent Capacitance

---

Formula

$$C=C_1+C_2+C_3$$

---

Solution

$$C = 5+8+10$$
$$C=23\mu F$$

---

Final Answer

$$\boxed{C=23\mu F}$$

---

Concept Behind the Numerical

In parallel:

• Potential difference remains the same.
• Equivalent capacitance is greater than the largest individual capacitor. 

  ---

  Board Exam Tip

  A quick way to remember:

  • Series → Reciprocal Formula
  • Parallel → Direct Addition

---

---

Solved Numerical 11.21

Zero Electric Field Location Between Two Charges

Difficulty Level: 🔴 Challenging

---

Problem

Two point charges, q₁​ = −1×10⁻⁶ C and q₂ ​= 4×10⁻⁶ C are separated by 3 m. Find the position where the resultant electric field is zero.

---

Given

$$q_1=-1\times10^{-6}C$$ $$q_2=4\times10^{-6}C$$
$$d=3m$$

---

Required

Location of zero electric field.

---

Formula

For equilibrium,

$$E_1=E_2$$
$$\frac{kq_1}{x^2} = \frac{kq_2}{(x-3)^2}$$

---

Solution

Since the charges are unlike, the zero-field point lies outside the two charges, on the side of the smaller charge.

Using

$$\frac{1}{x^2} = \frac{4}{(x-3)^2}$$

Solving,

$$x=-3m$$

---

Final Answer

$$\boxed{\text{The electric field is zero }3\,m\text{ to the left of }q_1.}$$

---

Concept Behind the Numerical

For unlike charges, the zero-field point always lies outside the charges and nearer the charge of smaller magnitude.

---

Board Exam Tip

Always draw a rough diagram before solving zero-field problems.

---

---

Solved Numerical 11.22

Capacitor with a Dielectric

Difficulty Level: 🟡 Moderate

---

Problem

A parallel plate capacitor has capacitance 5 μF. A dielectric of relative permittivity 4 completely fills the space between the plates. Calculate the new capacitance.

---

Given

$$C=5\mu F$$
$$K=4$$

---

Required

New capacitance.

---

Formula

$$C'=KC$$

---

Solution

$$C' = 4\times5$$
$$=20\mu F$$

---

Final Answer

$$\boxed{20\mu F}$$

---

Concept Behind the Numerical

A dielectric increases capacitance because it reduces the electric field between the plates.

---

Board Exam Tip

Capacitance always increases when a dielectric is inserted.

---

---

Solved Numerical 11.23

Mixed Capacitor Network

Difficulty Level: 🔴 Challenging

---

Problem

Two capacitors of 6 μF and 3 μF are connected in series. Their combination is connected in parallel with a 4 μF capacitor. Find the equivalent capacitance.

---

Given

$$C_1=6\mu F$$
$$C_2=3\mu F$$
$$C_3=4\mu F$$

---

Required

Equivalent capacitance.

---

Solution

Series combination

$$\frac1C = \frac16+\frac13$$
$$= \frac36$$
$$C=2\mu F$$

Now in parallel

$$C_{eq} = 2+4$$
$$=6\mu F$$

---

Final Answer

$$\boxed{6\mu F}$$

---

Concept Behind the Numerical

Always solve the series part first, then the parallel part.

---

Board Exam Tip

Work step by step. Never combine series and parallel simultaneously.

---

---

Solved Numerical 11.24

Energy Stored After Charging

Difficulty Level: 🟡 Moderate

---

Problem

A capacitor of 15 μF is charged to 300 V. Calculate the energy stored.

---

Given

$$C=15\mu F$$
$$V=300V$$

---

Formula

$$U=\frac12CV^2$$

---

Solution

$$U = \frac12 (15\times10^{-6}) (300)^2$$
$$= 0.675J$$

---

Final Answer

$$\boxed{0.675J}$$

---

Concept Behind the Numerical

Electrical energy is stored in the electric field between capacitor plates.

---

Board Exam Tip

When voltage is known,

always use

$$U=\frac12CV^2$$

---

---

Solved Numerical 11.25

Electric Field Between Parallel Plates

Difficulty Level: 🟢 Easy

---

Problem

Two parallel plates are 5 cm apart and connected to a 250 V battery.

Calculate the electric field strength between the plates.

---

Given

$$V=250V$$
$$d=5cm =0.05m$$

---

Formula

$$E=\frac{V}{d}$$

---

Solution

$$E = \frac{250}{0.05}$$
$$= 5000N/C$$

---

Final Answer

$$\boxed{5000N/C}$$

---

Concept Behind the Numerical

The electric field between parallel plates is uniform.

---

Board Exam Tip

Always convert centimetres into metres before substitution.

---

---

Solved Numerical 11.26

Electric Field Due to an Infinite Charged Sheet

Difficulty Level: 🔴 Challenging

---

Problem

An infinite plane sheet carries a uniform surface charge density of σ = 8.0×10⁻⁸ C/m². Calculate the electric field near the sheet. Take  εₒ ​= 8.85×10⁻¹² F/m.

---

Given

$$\sigma=8.0\times10^{-8}C/m^2$$
$$\varepsilon_0=8.85\times10^{-12}F/m$$

---

Required

Electric field intensity.

---

Formula

$$E=\frac{\sigma}{2\varepsilon_0}$$

---

Solution

Substitute the values:

$$E= \frac{8.0\times10^{-8}} {2(8.85\times10^{-12})}$$
$$E = 4.52\times10^3N/C$$

---

Final Answer

$$\boxed{E=4.52\times10^3N/C}$$

---

Concept Behind the Numerical

Unlike a point charge, the electric field due to an infinite sheet is uniform and does not depend on distance.

---

Board Exam Tip

For an infinite sheet,

$$E=\frac{\sigma}{2\varepsilon_0}$$

Distance never appears in the formula.

---

---

Solved Numerical 11.27

Application of Gauss's Law

Difficulty Level: 🔴 Challenging

---

Problem

A spherical Gaussian surface encloses a charge of 6×10⁻⁶ C. Calculate the electric flux through the surface. Take εₒ ​= 8.85×10⁻¹² F/m.

---

Given

$$Q=6\times10^{-6}C$$

---

Required

Electric flux.

---

Formula

$$\Phi=\frac{Q}{\varepsilon_0}$$

---

Solution

$$\Phi = \frac{6\times10^{-6}} {8.85\times10^{-12}}$$
$$\Phi = 6.78\times10^5Nm^2/C$$

---

Final Answer

$$\boxed{\Phi=6.78\times10^5Nm^2/C}$$

---

Concept Behind the Numerical

According to Gauss's Law, electric flux depends only on the enclosed charge, not on the size or shape of the Gaussian surface.

---

Board Exam Tip

Ignore the radius of the Gaussian surface unless specifically required.

---

---

Solved Numerical 11.28

Potential of an Isolated Conducting Sphere

Difficulty Level: 🟡 Moderate

---

Problem

A conducting sphere of radius 0.15m carries a charge of 3×10⁻⁸ C. Find its electric potential.

---

Given

$$Q=3\times10^{-8}C$$
$$R=0.15m$$

---

Required

Electric potential.

---

Formula

$$V=\frac{kQ}{R}$$

---

Solution

$$V = \frac{(9\times10^9)(3\times10^{-8})} {0.15}$$
$$V = 1800V$$

---

Final Answer

$$\boxed{V=1800V}$$

---

Concept Behind the Numerical

The entire conducting sphere remains at the same electric potential.

---

Board Exam Tip

Remember that the potential is the same at the surface and inside a charged conductor.

---

---

Solved Numerical 11.29

Combined Electric Field and Potential

Difficulty Level: 🔴 Challenging

---

Problem

A point charge of 5×10⁻⁶ C is located in free space. Calculate the:

(a) Electric field intensity

(b) Electric potential

at a point 0.50 m away.

---

Given

$$Q=5\times10^{-6}C$$
$$r=0.50m$$

---

Required

Electric field and electric potential.

---

Formula

Electric field

$$E=\frac{kQ}{r^2}$$

Electric potential

$$V=\frac{kQ}{r}$$

---

Solution

Electric Field

$$E = \frac{(9\times10^9)(5\times10^{-6})} {(0.50)^2}$$
$$E = 1.8\times10^5N/C$$

---

Electric Potential

$$V = \frac{(9\times10^9)(5\times10^{-6})} {0.50}$$
$$V = 9.0\times10^4V$$

---

Final Answer

Electric Field

$$\boxed{1.8\times10^5N/C}$$

Electric Potential

$$\boxed{9.0\times10^4V}$$

---

Concept Behind the Numerical

Although both depend on distance,

• Electric field varies as

$$\frac1{r^2}$$

• Electric potential varies as

$$\frac1r$$

---

Board Exam Tip

Never confuse the formulas for electric field and electric potential.

---

---

Solved Numerical 11.30

Comprehensive Electrostatics Challenge

Difficulty Level: 🔴 Challenging

---

Problem

A capacitor of capacitance 10μF is charged by a 400 V battery. 

Calculate:

(a) Charge stored

(b) Energy stored

(c) Number of electrons transferred

Take

$$e=1.60\times10^{-19}C$$

---

Given

$$C=10\times10^{-6}F$$
$$V=400V$$

---

Required

• Charge
• Energy
• Number of electrons

---

Formula

Charge

$$Q=CV$$

Energy

$$U=\frac12CV^2$$

Number of electrons

$$n=\frac{Q}{e}$$

---

Solution

(a) Charge

$$Q = (10\times10^{-6})(400)$$
$$Q = 4.0\times10^{-3}C$$

---

(b) Energy

$$U = \frac12 (10\times10^{-6}) (400)^2$$
$$U = 0.80J$$

---

(c) Number of Electrons

$$n = \frac{4.0\times10^{-3}} {1.60\times10^{-19}}$$
$$n = 2.5\times10^{16}$$

---

Final Answer

Charge

$$\boxed{4.0\times10^{-3}C}$$

Energy

$$\boxed{{\displaystyle 0.80J}}$$

Number of Electrons

$$\boxed{2.5\times10^{16}}$$

---

Concept Behind the Numerical 

This problem combines the three most important capacitor relationships:

• Charge stored
• Energy stored
• Electron transfer

---

Board Exam Tip

For capacitor problems, remember these three equations:

$$Q=CV$$
$$U=\frac12CV^2$$
$$n=\frac{Q}{e}$$

These are among the most frequently tested formulas in electrostatics.

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