Unit 11 - Electrostatics, Class 12 Physics – Important MCQs, Questions & Answers with Solved Numerical

Class 12 Physics Unit 11 - Electrostatics– Important MCQs, Questions & Answers with Solved Numerical

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Class 11 Physics – Unit-wise MCQs & Study Resources (Main-Page)

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This post covers all important topics of Unit-10: Thermodynamics for Class 11 Physics, including the zeroth, first, and second laws of thermodynamics, heat transfer, work, internal energy, specific and molar heats, Carnot engine, refrigerator, reversible and irreversible processes, entropy, and efficiency of heat engines.

Perfect for exam preparation, quick revision, board exams, and competitive test practice, this comprehensive set of questions, short answers, conceptual explanations, and numericals helps students strengthen their understanding of thermal physics and problem-solving skills.

Learn and master all key concepts with step-by-step numerical solutions, clear explanations, and concise conceptual answers, making it ideal for students aiming for high marks and complete mastery of Unit-10: Thermodynamics.

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Multiple Choice Question (MCQs) With answer and explanation (Class-12 Unit-11 Electrostatics)

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MCQs No. 1

A charge Q is divided into two parts q and (Q − q) and separated by a distance R. The force of repulsion between them will be maximum when:

a. q = Q/4
b. q = Q/2
c. q = Q
d. None of these

Correct answer: b. q = Q/2

Explanation:
Force F = k q(Q−q)/R². This product becomes maximum when q = Q/2, so force is maximum for equal division.

According to Coulomb’s law,

$$F = k \frac{q(Q-q)}{R^2}$$

Since k and R are constants, the force depends only on:

$$F \propto q(Q-q)$$

Let:

$$f(q) = q(Q-q) = Qq - q^2$$

To find maximum force, differentiate with respect to q:

$$\frac{df}{dq} = Q - 2q = 0$$

or

$$q = \frac{Q}{2}$$

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MCQs No. 2

Some charge is given to a conductor. Then its potential:

a. Is maximum at surface
b. Is maximum at centre
c. Remains same throughout the conductor
d. Is maximum somewhere between surface and centre

Correct answer: c. Remains same throughout the conductor

Explanation:
In electrostatic equilibrium, electric field inside a conductor is zero, so potential remains constant everywhere.

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MCQs No. 3

Electric potential of earth is taken to be zero because earth is a good:

a. Semiconductor
b. Conductor
c. Insulator
d. Dielectric

Correct answer: b. Conductor

Explanation:
Earth can accept or supply large amount of charge without changing its potential, so it is treated as zero potential.

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MCQs No. 4

A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1 kV, its kinetic energy will be:

a. 1840 keV
b. 1/1840 keV
c. 1 keV
d. 920 keV

Correct answer: c. 1 keV

Explanation:
Kinetic energy gained = qV. Since charge of proton equals electron, both gain same energy = 1 keV.

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MCQs No. 5

A capacitor is charged with a battery and then disconnected. A dielectric slab is inserted between the plates. Then:

a. Charge decreases and potential increases
b. Potential increases, energy decreases, charge same
c. Potential decreases, stored energy decreases, charge unchanged
d. None of these

Correct answer: c. Potential decreases, stored energy decreases, charge unchanged

Explanation:
After disconnection, charge remains constant. Dielectric increases capacitance, so potential and energy decrease.

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MCQs No. 6

A 1 μF capacitor is subjected to 4000 V. The energy stored is:

a. 8 J
b. 16 J
c. 4 × 10⁻³ J
d. 2 × 10⁻³ J

Correct answer: a. 8 J

Explanation:
U = ½CV² = ½ × 10⁻⁶ × (4000)² = 8 J.

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MCQs No. 7

(In given figure) the charge on 3 μF capacitor is:

a. 5 μC
b. 10 μC
c. 3 μC
d. 6 μC

Correct answer: b. 10 μC

Explanation:

Since all capacitors are connected in series, an important rule applies:

Charge on each capacitor in series is the same.

So,

$$Q_1=Q_2=Q_3=Q$$

To Find equivalent capacitance (Cₑ)

For series combination:

$$$$1Ce=12+13+16

Taking LCM = 6:

$$\frac{1}{C_e}=\frac{3+2+1}{6}=\frac{6}{6}=1$$$$C_e=1\mu F$$

To Find total charge

$$Q=C_{e}V$$$$Q=1\mu F\times 10V=10\mu C$$

Charge on 3 μF capacitor

Because capacitors are in series:

$$Q_{3\mu F}=Q=10\mu C$$

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MCQs No. 8

The electric potential difference between A and B is ΔV. Work done by field in moving charge q from A to B is:

a. W = −qΔV
b. W = qΔV
c. W = −ΔV/q
d. W = ΔV/q

Correct answer: b. W = qΔV

Explanation:
Work done by electric field equals charge × potential difference.

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MCQs No. 9

Electric flux through a spherical surface due to a charge at its centre depends on:

a. Radius of sphere
b. Charge outside sphere
c. Surface area
d. Charge inside sphere

Correct answer: d. Charge inside sphere

Explanation:
By Gauss’s law, Φ = Q/ε₀. Flux depends only on enclosed charge.

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MCQs No. 10

10. Two parallel, metal plates are a distance 8.00 m apart. The electric field between the plates is uniform, directed toward the right, and has a magnitude of 4.00 N/C. If an ion of charge+2e is released at rest at the left-hand plate, what is its kinetic energy when it reaches the right-hand plate?

a. 4 eV
b. 64 eV
c. 32 eV
d. 16 eV

Correct answer: b. 64 eV

Explanation:

To find potential difference between plates

$$V = E d$$

$$V = 4.00 \times 8.00 = 32 \text{ V}$$

Using relation for kinetic energy gained

$$KE = qV$$

Charge of ion:

$$q = 2e = 2(1.6\times10^{-19})\;C$$

But when energy is asked in electron-volts, simply use:

$$KE = (2e)(32V)=64\;eV$$

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SHORT QUESTIONS WITH BRIEF ANSWERS (Unit-7 Oscillation)

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Part A: Conceptual Questions

1. What is meant by a thermodynamic path?
A thermodynamic path is the sequence of states through which a system passes during a process. The work and heat depend on the path taken.

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2. Why is work done by a gas represented as area under a P–V curve?
Because work is given by $W = \int P\,dV$, and graphically this corresponds to the area under the pressure–volume diagram.

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3. Why does internal energy of an ideal gas depend only on temperature?
In an ideal gas, intermolecular forces are negligible, so internal energy depends only on kinetic energy of molecules, which is a function of temperature.

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4. What is the physical meaning of a quasi‑static process?
A quasi‑static process occurs infinitely slowly so that the system remains in equilibrium at every instant.

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5. Why can’t a perpetual motion machine of the first kind exist?
Because it would violate the first law of thermodynamics by producing work without energy input.

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6. Why can’t a perpetual motion machine of the second kind exist?
Because it would violate the second law of thermodynamics by converting all heat into work with no heat rejection.

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7. What happens to the entropy of an isolated system in a reversible process?
In a reversible process, the entropy of an isolated system remains constant.

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8. What is the zeroth law of thermodynamics used for?
It establishes the concept of temperature and allows the use of thermometers by defining thermal equilibrium.

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9. Why is heat considered a path function?
Because the amount of heat transferred depends on how the process is carried out and not only on the initial and final states.

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10. What is the difference between cyclic and non‑cyclic processes?
In a cyclic process, the system returns to its initial state at the end of the cycle; in non‑cyclic, it does not.

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11. Why does an adiabatic process result in a temperature change?
Because no heat enters or leaves the system, so work done changes the internal energy, changing the temperature.

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12. In which direction does heat flow spontaneously?
Heat flows spontaneously from a region of higher temperature to a region of lower temperature.

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13. Why does increasing temperature increase entropy?
Higher temperature means more random molecular motion, so disorder and entropy increase.

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14. Why does an increase in pressure at constant temperature decrease entropy?
Increasing pressure at constant temperature reduces volume and randomness, thus reducing entropy.

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15. What is the basic purpose of a heat pump?
A heat pump transfers heat from a cold region to a hot region using work, often for space heating.

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16. Why is the efficiency of real heat engines lower than Carnot efficiency?
Because of irreversibilities like friction, heat losses, and non‑ideal processes that reduce work output.

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17. What does the term “thermal reservoir” mean?
A thermal reservoir is a body large enough that its temperature remains essentially constant even when it gives or absorbs heat.

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18. Why is specific heat capacity different at constant pressure and constant volume?
At constant pressure, heat must raise temperature and do expansion work, so $C_p > C_v$.

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19. What does the symbol $\gamma$ represent for a gas?
$\gamma = C_p / C_v$, called the adiabatic index.

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20. Why is entropy considered a measure of disorder?

Because it quantifies the number of microscopic arrangements corresponding to a macroscopic state — more arrangements mean more disorder..

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LONG / COMPREHENSIVE QUESTIONS WITH BRIEF EXAM ANSWERS

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1. Define a thermodynamic system, surroundings, and boundary. Explain state functions and path functions.

Answer:
A thermodynamic system is the part of the universe under study, while everything else is the surroundings. The boundary separates the system from the surroundings and can be fixed or movable. State functions (like pressure, volume, temperature, internal energy) depend only on the state of the system, not on the path. Path functions (like heat and work) depend on the specific process taken.

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2. Distinguish between heat, work, and internal energy.

Answer:
Heat (Q) is energy transferred due to temperature difference.
Work (W) is energy transfer due to force causing displacement.
Internal energy (U) is the total microscopic energy of the system, including kinetic and potential energies of molecules. Heat and work are modes of energy transfer, while internal energy is a state property.

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3. State and explain the first law of thermodynamics.

Answer:
The first law states that energy cannot be created or destroyed, only transformed:

$$\Delta U = Q - W$$

Here, $\Delta U$ is the change in internal energy, $Q$ is heat added, and $W$ is work done by the system. This law expresses conservation of energy, indicating heat can be converted into work or internal energy, but total energy remains constant.

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4. Define Cp and Cv. Show that Cp − Cv = R for an ideal gas.

Answer:
Cv is the molar heat capacity at constant volume; Cp is at constant pressure.
For one mole of ideal gas, using first law:

$$Q = \Delta U + W$$

At constant volume, $W = 0 \Rightarrow Q = \Delta U = Cv \Delta T$ 

At constant pressure, $W = P\Delta V = R\Delta T \Rightarrow Cp \Delta T = Cv \Delta T + R \Delta T \Rightarrow Cp - Cv = R$

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5. Explain reversible and irreversible processes with examples.

Answer:
A reversible process occurs infinitely slowly and can return to its initial state without leaving any changes in system or surroundings (e.g., slow isothermal expansion).
An irreversible process occurs naturally and cannot be reversed without changes (e.g., free expansion, friction, spontaneous heat flow).

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6. What is a heat engine? Explain its efficiency.

Answer:
A heat engine converts heat from a high-temperature reservoir into mechanical work and rejects some heat to a low-temperature reservoir.
Efficiency:

$$\eta = \frac{W}{Q_h} = 1 - \frac{Q_c}{Q_h}$$

where $Q_h$  is heat absorbed and $Q_c$ is heat rejected. Maximum efficiency depends on temperatures of reservoirs, not the working substance.

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7. State and explain the second law of thermodynamics.

Answer:
Kelvin-Planck statement: No engine can convert all absorbed heat into work without rejecting some heat.
Clausius statement: Heat cannot flow spontaneously from a colder body to a hotter body.
The second law introduces directionality of natural processes and entropy, explaining why some energy transformations are irreversible.

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8. Explain Carnot engine and its efficiency.

Answer:
A Carnot engine is an ideal reversible engine operating between two reservoirs. Its efficiency:

$$\eta = 1 - \frac{T_c}{T_h}$$

It represents the maximum possible efficiency between temperatures $T_h$ and $T_c$ . All reversible engines between the same reservoirs have the same efficiency.

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9. What is a refrigerator? Derive the coefficient of performance (COP).

Answer:
A refrigerator removes heat from a cold reservoir and rejects it to a hot reservoir using work.

$$COP = \frac{Q_c}{W} = \frac{Q_c}{Q_h - Q_c} \quad \text{(ideal)}$$

For Carnot refrigerator:

$$COP = \frac{T_c}{T_h - T_c}$$

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10. Explain entropy and its significance. How does it relate to the second law?

Answer:
Entropy (S) measures the disorder or randomness of a system:

$$dS = \frac{dQ_{rev}}{T}$$Entropy increases in irreversible processes, remains constant in reversible adiabatic processes, and is a state function. The second law in terms of entropy: “Entropy of the universe always increases for natural processes.” This explains energy degradation and direction of processes. 

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Miscellaneous Numerical Problems with Proper Solutions

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Numerical 1

Problem:
One mole of an ideal gas initially at 300 K and 1 atm is compressed isothermally to half its original volume. Calculate the work done by the gas. (R = 8.314 J/mol·K)

Given:
n=1 mol, T=300 K, V2=1/2 V1, R=8.314 J/mol.K

To Find: Work $W$

Solution:
Isothermal work:

$$W = nRT \ln\frac{V_2}{V_1} = 1(8.314)(300)\ln\left(\frac{1}{2}\right) = 2494.2(-0.693) \approx -1729\,\text{J}$$

Answer:
$W \approx -1729\,\text{J}$ (work done on the gas)

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Numerical 2

Problem:
A gas absorbs 1200 J of heat and does 450 J of work. Calculate the change in internal energy.

Given:
$Q=1200\,\text{J}, W=450\,\text{J}$ 

To Find: $\Delta U$

Solution:
First Law:

$$\Delta U = Q - W = 1200 - 450 = 750\,\text{J}$$

Answer:
$\Delta U = 750\,\text{J}$ 

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Numerical 3

Problem:
An ideal gas expands at constant pressure of $2 \times 10^5\ \text{Pa}$ from 2 L to 6 L.
(a) Calculate the work done by the gas.
(b) If the heat supplied is 1500 J, find the change in internal energy.

Given:
$P=2\times 10^5\,\text{Pa}, V_1=2\,\text{L}, V_2=6\,\text{L}, Q=1500\,\text{J}$ 

To Find: $W, \Delta U$ 

Solution:
Convert L to m³:
$V_1=2\times10^{-3}, V_2=6\times10^{-3}\,\text{m³}$ 

(a)

$$W = P\Delta V = 2\times10^5(6-2)\times10^{-3} = 2\times10^5(4\times10^{-3}) = 800\,\text{J}$$

(b) First law:

$$\Delta U = Q - W = 1500-800 = 700\,\text{J}$$

Answer:
$W = 800\,\text{J}, \Delta U = 700\,\text{J}$ 

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Numerical 4

Problem:
A Carnot engine works between 727 K and 327 K. Find the efficiency of the engine.

Given:
$T_h=727\,\text{K}, T_c=327\,\text{K}$ 

To Find: Efficiency $\eta$

Solution:

$$\eta = 1 - \frac{T_c}{T_h} = 1 - \frac{327}{727} \approx 1 - 0.45 = 0.55$$

Answer:
$\eta \approx 55\%$

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Numerical 5

Problem:
A Carnot refrigerator operates between 280 K and 300 K. If it removes 800 J of heat from the cold reservoir each cycle, calculate the work done per cycle.

Given:
$T_c=280\,\text{K}, T_h=300\,\text{K}, Q_c=800\,\text{J}$ 

To Find: Work $W$

Solution:
COP:

$$COP = \frac{T_c}{T_h - T_c} = \frac{280}{20} = 14$$$$W = \frac{Q_c}{COP} = \frac{800}{14} \approx 57.1\,\text{J}$$

Answer:
$W \approx 57.1\,\text{J}$

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Numerical 6

Problem:
Calculate the change in entropy when 4 kg of water at 50 °C is heated to 90 °C. Specific heat $c=4180\ \text{J/kg·K}$.

Given:

$m=4\,\text{kg}, c=4180\,\text{J/kg·K}, T_1=323\,\text{K}, T_2=363\,\text{K}$

To Find: $\Delta S$

Solution:

$$\Delta S = mc\ln\frac{T_2}{T_1} = 4(4180)\ln\frac{363}{323}$$$$\ln\frac{363}{323} \approx \ln(1.124) \approx 0.117$$$$\Delta S = 16720(0.117) \approx 1956\,\text{J/K}$$

Answer:
$\Delta S\approx 1956\text{J/K}$

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Numerical 7

Problem:
An ideal gas at 27 °C and 2 atm occupies 4 L. Calculate the number of moles present.

Given:
$T=300\,\text{K}, P=2\times1.013\times10^5=2.026\times10^5\ \text{Pa}, V=4\times10^{-3}\,\text{m³}, R=8.314$

To Find: $n$

Solution:
Ideal gas law:

$$n = \frac{PV}{RT} = \frac{(2.026\times10^5)(4\times10^{-3})}{8.314(300)}$$$$= \frac{810.4}{2494.2} \approx 0.325\,\text{mol}$$

Answer:
$n \approx 0.325\,\text{mol}$

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Numerical 8

Problem:
During an adiabatic process, the temperature of a gas falls from 400 K to 320 K. If γ=1.4, find the ratio of the final to initial volume $(V_2/V_1)$.

Given:
$T_1=400\,\text{K}, T_2=320\,\text{K}, \gamma=1.4$ 

To Find: $V_2/V_1$

Solution:
For adiabatic:

$$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$$$$\frac{V_2}{V_1}=\left(\frac{T_1}{T_2}\right)^{1/(\gamma-1)}=\left(\frac{400}{320}\right)^{1/0.4}$$$$= (1.25)^{2.5} \approx 1.25^{2.5} \approx 1.25^2(1.25^{0.5})\approx 1.56(1.118) \approx 1.744$$

Answer:
$V_2/V_1 \approx 1.74$ 

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Numerical 9

Problem:
A gas at 300 K expands from 2 L to 8 L at constant pressure. Calculate the work done. (1 L = $10^{-3}$m³)

Given:

$P=1\times10^5\,\text{Pa}, V_1=2\times10^{-3}, V_2=8\times10^{-3}$ 

To Find:  $W$

Solution:

$$W=P\Delta V=1\times10^5(8-2)\times10^{-3}=1\times10^5(6\times10^{-3})=600\,\text{J}$$

Answer:
$W=600\text{J}$

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Numerical 10

Problem:
A gas absorbs 900 J of heat and 400 J of work is done on the gas. Find the change in internal energy.

Given:

$Q=900\,\text{J}, W_{\text{on gas}}=400\,\text{J}$

To Find: $\Delta U$

Solution:
First Law (work done on gas is negative in sign convention of $\Delta U = Q - W$):

$$\Delta U = Q + W_{\text{on}}=900+400=1300\,\text{J}$$

Answer:
$\Delta U=1300\,\text{J}$ 

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Click on the link below for complete Class 11 Physics notes, MCQs, Q&A, and Numericals.

Class 11 Physics – Unit-wise MCQs & Study Resources (Main-Page)

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MCQs No. 7

In the given circuit, three capacitors 2 μF, 3 μF, and 6 μF are connected in series across a 10 V battery. Find the charge on the 3 μF capacitor.

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Explanation:

Since all capacitors are connected in series, an important rule applies:

Charge on each capacitor in series is the same.

So,

$$Q_1 = Q_2 = Q_3 = Q$$

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To Find equivalent capacitance (Cₑ)

For series combination:

$$\frac{1}{C_e}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$$

Taking LCM = 6:

$$\frac{1}{C_e}=\frac{3+2+1}{6}=\frac{6}{6}=1$$$$C_e=1\,\mu F$$

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To Find total charge

$$Q=C_e V$$$$Q=1\,\mu F \times 10\,V=10\,\mu C$$

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Charge on 3 μF capacitor

Because capacitors are in series:

$$Q_{3\mu F}=Q=10\,\mu C$$