Two capacitors C1 and C2 are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are U1 and U2 , respectively. Which of the given statements is true?

Since both capacitors are connected in parallel across the same battery, their potential difference V is identical. From the given charge-time graph, at any specific instant of time, the charge accumulated on the second capacitor is greater than that on the first: Using the relation q = CV, since V is identical, we get: Now, the electrostatic energy stored in a capacitor is given by: Since C2 > C1 and V is constant, the stored energy satisfies: Thus, the correct relationships are C2 > C1 and U2 > U1. Sees parallel connection → Immediately lock potential difference V as constant. This simplifies q propto C and U propto C, creating a direct linear bridge from graph height to energy capacity. Class 12 Physics: Electrostatic Potential and Capacitance

Two capacitors C1 and C2 are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are U1 and U2 , respectively. Which of the given s

Since both capacitors are connected in parallel across the same battery, their potential difference V is identical. From the given charge-time graph, at any specific instant of time, the charge accumulated on the second capacitor is greater than that on the first: Using the relation q = CV, since V is identical, we get: Now, the electrostatic energy stored in a capacitor is given by: Since C2 > C1 and V is constant, the stored energy satisfies: Thus, the correct relationships are C2 > C1 and U2

Two capacitors C1 and C2 are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are U1 and U2 , respectively. Which of the given statements is true?

Since both capacitors are connected in parallel across the same battery, their potential difference V is identical. From the given charge-time graph, at any specific instant of time, the charge accumulated on the second capacitor is greater than that on the first: Using the relation q = CV, since V is identical, we get: Now, the electrostatic energy stored in a capacitor is given by: Since C2 > C1 and V is constant, the stored energy satisfies: Thus, the correct relationships are C2 > C1 and U2 > U1. Sees parallel connection → Immediately lock potential difference V as constant. This simplifies q propto C and U propto C, creating a direct linear bridge from graph height to energy capacity. Class 12 Physics: Electrostatic Potential and Capacitance

Two capacitors C1 and C2 are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are U1 and U2 , respectively. Which of the given s

Since both capacitors are connected in parallel across the same battery, their potential difference V is identical. From the given charge-time graph, at any specific instant of time, the charge accumulated on the second capacitor is greater than that on the first: Using the relation q = CV, since V is identical, we get: Now, the electrostatic energy stored in a capacitor is given by: Since C2 > C1 and V is constant, the stored energy satisfies: Thus, the correct relationships are C2 > C1 and U2

Two capacitors C1 and C2 are connected in parallel to a battery. Charge-time graph is shown below for the two…

Year	2026	2025	2024	Total
Questions	3	7	3	13

More Electrostatic Potential and Capacitance Previous-Year Questions — Page 2

Q jee_main_2025_07_april_morning Electrostatic Potential Energy

Two charges

q₁

and

q₂

are separated by a distance of 30 cm. A third charge

q₃

initially at 'C' as shown in the figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to movement of

q₃

from C to D is given by

(q₃ K)/(4π ε₀)

, the value of K is:

Electrostatic configuration with circular path for Q20 - JEE Main 2025 Morning — A charges layout with q1 at the center of the arc, q2 at 30 cm from q1, and q3 moving along the circular boundary of radius 40 cm.

A. $8 q₂$ $8 \mathrm{q}_{2}$
B. $6 q₂$ $6 \mathrm{q}_{2}$
C. $8q₁$ $8\mathrm{q}_{1}$
D. $6 q₁$ $6 \mathrm{q}_{1}$

Solution

Related Formula

The change in potential energy $Δ U$ $\Delta U$ of a charge $q$ $q$ moved between two points of potentials $VC$ $V_C$ and $VD$ $V_D$ is:

Δ U = q₃ (VD - VC)

Potential due to a point charge $q$ $q$ at distance $r$ $r$ is:

V = (1)/(4πε₀) (q)/(r)

Core Logic

Let point $A$ $A$ hold charge $q₁$ $q_1$ and $B$ $B$ hold $q₂$ $q_2$ separated by $30 ~cm = 0.3 ~m$ $30 \mathrm{~cm} = 0.3 \mathrm{~m}$.

The path of $q₃$ $q_3$ is a circular arc of radius $R = 40 ~cm = 0.4 ~m$ $R = 40 \mathrm{~cm} = 0.4 \mathrm{~m}$ centered at $A$ $A$. Thus, distance of $C$ $C$ and $D$ $D$ from $q₁$ $q_1$ is constant:

r1C = r1D = 0.4 ~m

Distance of $C$ $C$ from $q₂$ $q_2$ ( $B$ $B$):

r2C = \sqrt(AC² + AB²) = \sqrt(40² + 30²) = 50 ~cm = 0.5 ~m

Distance of $D$ $D$ from $q₂$ $q_2$ ( $B$ $B$):

r2D = AD - AB = 40 cm - 30 cm = 10 cm = 0.1 m

Step 1: Calculate Potentials

Potential at $C$ $C$ due to $q₁$ $q_1$ and $q₂$ $q_2$:

VC = (1)/(4πε₀) ( (q₁)/(0.4) + (q₂)/(0.5) )

Potential at $D$ $D$ due to $q₁$ $q_1$ and $q₂$ $q_2$:

VD = (1)/(4πε₀) ( (q₁)/(0.4) + (q₂)/(0.1) )

Step 2: Difference in Potential Energy

The potential difference is:

VD - VC = (1)/(4πε₀) ( (q₂)/(0.1) - (q₂)/(0.5) ) = (q₂)/(4πε₀) [10 - 2] = (8q₂)/(4πε₀)

(Notice that the potential contribution of $q₁$ $q_1$ cancels out because $C$ $C$ and $D$ $D$ are equidistant from $q₁$ $q_1$).

The change in potential energy is:

Δ U = q₃ (VD - VC) = (q₃ (8q₂))/(4πε₀)

Comparing with $(q₃ K)/(4πε₀)$ $\frac{q_3 K}{4\pi\epsilon_0}$ yields $K = 8q₂$ $K = 8q_2$.

Pattern Recognition

Sees: Arc path centered on one of the charges. Shortcut: Since the path is circular about $q₁$ $q_1$, $q₁$ $q_1$ contributes nothing to the potential difference between the end points. The entire change in potential energy is due to $q₂$ $q_2$. The distance difference translates to potential difference $Δ V = q₂ ((1)/(0.1) - (1)/(0.5)) = 8q₂$ , so $K = 8q₂$ $K = 8q_2$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q14 jee_main_2025_29_jan_evening Sharing of Charges and Loss of Energy

A capacitor,

C₁ = 6

is charged to a potential difference of

V₀ = 5V

using a

5V

battery. The battery is removed and another capacitor,

C₂ = 12

is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges

(q₁

and

q₂)

on the capacitors

C₁

and

C₂

when equilibrium condition is reached.

Sharing of Charges and Loss of Energy diagram for Q14 - JEE Main 2025 Evening — The circuit diagram displays two capacitors C1 and C2 with a switch S used to establish parallel connectivity between them.

A. $q₁ = 15 , q₂ = 30$
B. $q₁ = 30 , q₂ = 15$
C. $q₁ = 10 , q₂ = 20$
D. $q₁ = 20 , q₂ = 10$

Solution

Related Formula

Qtotal = C₁ V₀

Vc = QtotalC₁ + C₂

$q = C \cdot Vc$ $q = C \cdot V_c$

Core Logic

Initial Charge Calculation:

Before closing the switch, capacitor $C₁$ $C_1$ accumulates total charge:

q₁' = C₁ \cdot V₀ = 6 \times 5V = 30

Uncharged capacitor $C₂$ $C_2$ holds $q₂' = 0$ $q_2' = 0$.

Sharing of Charges Initial State diagram for Q14 - JEE Main 2025 Evening — The circuit diagram displays two capacitors C1 and C2 with a switch S used to establish parallel connectivity between them.

Redistribution at Equilibrium:

Closing the switch causes parallel redistribution until they reach a common potential $Vc$ $V_c$:

Vc = 30 + 06 + 12 = (30)/(18) = (5)/(3)~V

Final Charges:

q₁ = C₁ \cdot Vc = 6 \times (5)/(3) = 10

q₂ = C₂ \cdot Vc = 12 \times (5)/(3) = 20

Hence, the charges are $10$ $10\mu\mathrm{C}$ and $20$ $20\mu\mathrm{C}$ respectively.

Pattern Recognition

For parallel combination setups, the final charge splits in the exact direct ratio of their capacitances: $q₁ : q₂ = C₁ : C₂ = 6 : 12 = 1 : 2$ . Out of $30$ $30\mu\mathrm{C}$, the breakdown must cleanly be $10$ $10\mu\mathrm{C}$ and $20$ $20\mu\mathrm{C}$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q4 jee_main_2025_28_jan_morning Equipotential Surfaces

Three infinitely long wires with linear charge density

λ

are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?

A. $xy + yz + zx = constant$
B. $(x + y)(y + z)(z + x) = constant$
C. $(x² +y²)(y² +z²)(z² +x²) = constant$
D. $xyz = constant$ $\mathrm{xyz} = \text{constant}$

Solution

Related Formula

V = -\int E r = 2kλ ln r + c

Core Logic

The potential at a point due to a line charge on an axis is proportional to the logarithm of its perpendicular distance.

For the wire along the z-axis: $Vz = -kλ ln(x² + y²)$

For the wire along the x-axis: $Vₓ = -kλ ln(y² + z²)$

For the wire along the y-axis: $Vy = -kλ ln(z² + x²)$

Summing the individual potentials to find the net configuration potential:

Vₙₑₜ = -kλ [ ln(x²+y²) + ln(y²+z²) + ln(z²+x²) ] + C'

Vₙₑₜ = -kλ ln [ (x²+y²)(y²+z²)(z²+x²) ] + C'

For an equipotential surface, set $Vₙₑₜ = constant$ :

Step 1: Final Expression

(x² + y²)(y² + z²)(z² + x²) = constant

This maps perfectly to option (3).

Pattern Recognition

Logarithmic combination rules transform scalar potential additions into products inside the functional argument: $Σ ln(rᵢ²) = ln(Π rᵢ²)$ .

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q7 jee_main_2025_24_jan_evening Electrostatic Potential Energy

In the first configuration (1) as shown in the figure

Electrostatic potential energy configuration 1 square charges Q7 — The diagram displays configuration 1 with 4 charges on corners and configuration 2 with charges on the midpoints of the square sides.

, four identical charges

(q₀)

are kept at the corners A, B, C and D of square of side length 'a'. In the second configuration (2)

, the same charges are shifted to mid points G, E, H and F, of the square. If

K= 14πε₀

, the difference between the potential energies of configuration (2) and (1) is given by:

A. $Kq₀²a(4\sqrt(2)-2)$
B. $Kq₀²a(3-\sqrt(2))$
C. $Kq₀²a(4-2\sqrt(2))$
D. $Kq₀²a(3\sqrt(2)-2)$

Solution

Related Formula

U = Σi < j K qᵢ qⱼrᵢⱼ

Core Logic

For configuration (1) with side length $a$ $a$:

4 pairs of adjacent side-charges with distance $a$ $a$.
2 pairs of diagonal charges with distance $\sqrt(2)a$ $\sqrt{2}a$.

U₁ = 4 \cdot (Kq₀²)/(a) + 2 \cdot Kq₀²\sqrt(2)a = (Kq₀²)/(a)(4 + \sqrt(2))

For configuration (2), charges lie on midpoints forming an inner square of side length $a' = a\sqrt(2)$ $a' = \frac{a}{\sqrt{2}}$:

4 pairs at distance $a\sqrt(2)$ $\frac{a}{\sqrt{2}}$.
2 pairs at diagonal distance $a$ $a$.

U₂ = 4 \cdot Kq₀² a\sqrt(2) + 2 \cdot (Kq₀²)/(a) = (Kq₀²)/(a)(4\sqrt(2) + 2)

Step 1: Finding the difference

U₂ - U₁ = (Kq₀²)/(a)[(4\sqrt(2) + 2) - (4 + \sqrt(2))] = (Kq₀²)/(a)(3\sqrt(2) - 2)

Pattern Recognition

The side of the new square formed by midpoints is scaled down by $1/\sqrt(2)$ $1/\sqrt{2}$. Therefore, interaction terms scale accordingly based on system geometric dimensions.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrostatic Potential and Capacitance Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Calculate Potentials

Step 2: Difference in Potential Energy

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Expression

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding the difference

Pattern Recognition

Chapter Mix

More Electrostatic Potential and Capacitance Questions — jee_main_2025_28_jan_morning

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