Three infinitely long wires with linear charge density lambda are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?

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 = -klambda (x2 + y2) For the wire along the x-axis: Vx = -klambda (y2 + z2) For the wire along the y-axis: Vy = -klambda (z2 + x2) Summing the individual potentials to find the net configuration potential: For an equipotential surface, set Vnet = constant: This maps perfectly to option (3). Logarithmic combination rules transform scalar potential additions into products inside the functional argument: (ri2) = ( ri2). Class 12 Physics: Electrostatic Potential and Capacitance

Three infinitely long wires with linear charge density lambda are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?

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 = -klambda (x2 + y2) For the wire along the x-axis: Vx = -klambda (y2 + z2) For the wire along the y-axis: Vy = -klambda (z2 + x2) Summing the individual potentials to find the net configuration potential: For an equipotential surface, set Vnet = constant: This maps perfectly to option (3). Logarithmic combination rules transform scalar potent

Three infinitely long wires with linear charge density lambda are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?

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 = -klambda (x2 + y2) For the wire along the x-axis: Vx = -klambda (y2 + z2) For the wire along the y-axis: Vy = -klambda (z2 + x2) Summing the individual potentials to find the net configuration potential: For an equipotential surface, set Vnet = constant: This maps perfectly to option (3). Logarithmic combination rules transform scalar potential additions into products inside the functional argument: (ri2) = ( ri2). Class 12 Physics: Electrostatic Potential and Capacitance

Three infinitely long wires with linear charge density lambda are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?

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 = -klambda (x2 + y2) For the wire along the x-axis: Vx = -klambda (y2 + z2) For the wire along the y-axis: Vy = -klambda (z2 + x2) Summing the individual potentials to find the net configuration potential: For an equipotential surface, set Vnet = constant: This maps perfectly to option (3). Logarithmic combination rules transform scalar potent

Three infinitely long wires with linear charge density lambda are placed along the x-axis, y-axis and z-axis…

Year	2026	2025	2024	Total
Questions	3	7	3	13

More Electrostatic Potential and Capacitance Previous-Year Questions — Page 3

Q35 jee_main_2024_01_february_morning Capacitors and Capacitance

Two identical capacitors have same capacitance

C

. One of them is charged to the potential

V

and other to the potential

2V

. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is:

A. $(1)/(4) CV²$ $\frac{1}{4} CV^2$
B. $2 CV²$ $2 CV^2$
C. $(1)/(2) CV²$ $\frac{1}{2} CV^2$
D. $(3)/(4) CV²$ $\frac{3}{4} CV^2$

Solution

Related Formula

Energy stored in a capacitor:

U = (1)/(2)CV²

Common potential after sharing charges:

VC = (q₁ + q₂)/(C₁ + C₂)

Loss of energy:

Δ U = (1)/(2) (C₁ C₂)/(C₁ + C₂) (V₁ - V₂)²

Core Logic

Initial energy of the individual systems:

Uᵢ = (1)/(2)CV² + (1)/(2)C(2V)² = (1)/(2)CV² + 2CV² = (5)/(2)CV²

Common potential $VC$ $V_C$ when joined in parallel:

VC = (CV + C(2V))/(C + C) = (3CV)/(2C) = (3)/(2)V

Step 1: Calculate Final Energy and Loss

Final energy of the combined system:

Uf = (1)/(2)(2C)VC² = C ((3)/(2)V)² = (9)/(4)CV²

Decrease in energy (loss):

Δ U = Uᵢ - Uf = (5)/(2)CV² - (9)/(4)CV² = (1)/(4)CV²

Pattern Recognition

Direct Formula Shortcut:

Δ U = (1)/(2) (C \cdot C)/(2C) (2V - V)² = (1)/(4)C(V)² = (1)/(4)CV²

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q56 jee_main_2024_29_january_evening Capacitance and Resistor Networks at Steady State

In the given figure, the charge stored in

6

capacitor, when points A and B are joined by a connecting wire is ________

Capacitor-resistor bridge circuit for Q56 - JEE Main 2024 29 January Shift 2 — The diagram displays a bridge-like network containing a 6 Ohm resistor, 3uF capacitor, 6uF capacitor, and 3 Ohm resistor powered by a 9V supply.

Numerical Answer. Answer: 36 to 36

Solution

Related Formula

At steady state, a capacitor acts as an open circuit to DC current. The charge on a capacitor is:

$Q = C Δ V$ $Q = C \Delta V$

Core Logic

When node $A$ $A$ and node $B$ $B$ are connected by a wire, they reach the same electrical potential ( $VA = VB = VM$ $V_A = V_B = V_M$).

At DC steady state, capacitors block current, so current only flows through the resistors from the $9 V$ $9\text{ V}$ source to ground:

The $6 Ω$ $6\ \Omega$ resistor is connected between $9 V$ $9\text{ V}$ and node $M$ $M$.
The $3 Ω$ $3\ \Omega$ resistor is connected between node $M$ $M$ and ground.

Hence, the equivalent series resistance for the DC current path is:

Req = 6 + 3 = 9 Ω

The steady state current is:

I = VReq = 9 V9 Ω = 1 A

Simplified equivalent steady-state circuit for Q56 — The diagram displays a bridge-like network containing a 6 Ohm resistor, 3uF capacitor, 6uF capacitor, and 3 Ohm resistor powered by a 9V supply.

Step 1: Calculate the Potential Difference and Charge

The potential at node $M$ $M$ ( $VM$ $V_M$) is:

VM = I \times 3 Ω = 1 A \times 3 Ω = 3 V

The $6$ $6\mu\text{F}$ capacitor is connected between the $9 V$ $9\text{ V}$ source and node $B$ $B$ (which is at potential $VM = 3 V$ $V_M = 3\text{ V}$).

Thus, the potential difference across the $6$ $6\mu\text{F}$ capacitor is:

Δ V₆ = 9 V - 3 V = 6 V

The charge stored in the $6$ $6\mu\text{F}$ capacitor is:

Q = C Δ V = 6 \times 6 V = 36

Thus, the charge stored is $36$ $36\ \mu\text{C}$.

Pattern Recognition

Shorting $A$ $A$ and $B$ $B$ makes the network a simple voltage divider for resistors at steady state. Once potential of the middle node is found ( $3 V$ $3\text{ V}$), the capacitor charge is calculated instantly using $Q = C Δ V$ $Q = C \Delta V$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q57 jee_main_2024_30_jan_morning Energy Loss in Capacitors

A capacitor of capacitance C and potential V has energy E. It is connected to another capacitor of capacitance 2C and potential 2V. Then the loss of energy is

(x)/(3)E

, where x is ________.

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

E = (1)/(2) C V²

Δ E = (1)/(2) (C₁ C₂)/(C₁ + C₂) (V₁ - V₂)²

Core Logic

When two charged capacitors are connected in parallel, charge redistributes until they reach a common potential. During this redistribution, energy is dissipated as heat, strictly governed by the standard loss formula.

Step 1: Assign Values

Initial energy of first capacitor: $E = (1)/(2) C V²$ $E = \frac{1}{2} C V^2$ Capacitor 1: $C₁ = C$ $C_1 = C$, $V₁ = V$ $V_1 = V$ Capacitor 2: $C₂ = 2C$ $C_2 = 2C$, $V₂ = 2V$ $V_2 = 2V$

Step 2: Calculate Energy Loss

Δ E = (1)/(2) ((C)(2C))/(C + 2C) (V - 2V)²

Δ E = (1)/(2) (2C²)/(3C) (-V)²

Δ E = (1)/(2) ((2C)/(3)) V²

Δ E = (2)/(3) ((1)/(2) C V²)

Δ E = (2)/(3) E

Step 3: Match the Pattern

Given loss is $(x)/(3)E$ $\frac{x}{3}E$. Comparing, we get $x = 2$ $x = 2$.

Pattern Recognition

The loss formula $Δ E = (1)/(2) Ceq(Δ V)²$ elegantly bypasses recalculating common potential $Vc$ $V_c$ and summing final state energies.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Three infinitely long wires with linear charge density λ$\lambda$ are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Expression

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrostatic Potential and Capacitance Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Calculate Final Energy and Loss

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the Potential Difference and Charge

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Assign Values

Step 2: Calculate Energy Loss

Step 3: Match the Pattern

Pattern Recognition

Chapter Mix

More Electrostatic Potential and Capacitance Questions — jee_main_2025_28_jan_morning

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Three infinitely long wires with linear charge density $λ$ $\lambda$ are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?