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

Q30 jee_main_2026_22_january_evening Electric Potential of Coalescing Bubbles

Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :

A. $1:31/3$ $1:3^{1/3}$
B. $1:22/3$ $1:2^{2/3}$
C. $32/3:1$ $3^{2/3}:1$
D. $1:32/3$ $1:3^{2/3}$

Solution

Related Formula

V = (kq)/(r)

Volume Conservation: N \cdot ((4)/(3)π r³) = (4)/(3)π R³

Core Logic

From volume conservation of 3 coalescing droplets:

3 ((4)/(3)π r³) = (4)/(3)π R³ R = 31/3r

Total charge on resultant bigger bubble $Q = 3q$ $Q = 3q$.

Calculating initial potential $Vᵢ$ $V_i$ and final potential $Vf$ $V_f$:

Vᵢ = (kq)/(r)

Vf = (k(3q))/(R) = 3kq31/3r = 32/3 (kq)/(r)

Ratio of initial to final potential:

(Vᵢ)/(Vf) = 132/3 = 1 : 32/3

Coalescing charged bubbles diagram for Q30 - JEE Main 2026 Evening

Step 1: Final Conclusion

The ratio of potentials is $1 : 32/3$ $1 : 3^{2/3}$.

Pattern Recognition

Coalescing droplets rule: For $N$ $N$ identical drops, $R = N1/3r$ $R = N^{1/3}r$ and $Q = Nq$ $Q = Nq$. Potential ratio $Vᵢ / Vf = 1 / N2/3$ $V_i / V_f = 1 / N^{2/3}$. For $N=3$ $N=3$, ratio is $1 / 32/3$ $1 / 3^{2/3}$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q48 jee_main_2026_22_january_evening Sharing of Charges between Capacitors

A capacitor P with capacitance

10 \times 10⁻⁶

F is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor P is connected across another capacitor Q with capacitance

20 \times 10⁻⁶

F. The charge on capacitor Q when equilibrium is established will be

α \times 10⁻⁵

C (assume capacitor Q does not have any charge initially), the value of

α

is ____.

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

Vcommon = (C₁ V₁ + C₂ V₂)/(C₁ + C₂)

Q₂ = C₂ Vcommon

Core Logic

Given $C₁ = 10 \times 10⁻⁶ ~F, V₁ = 6.0 ~V$ and $C₂ = 20 \times 10⁻⁶ ~F, V₂ = 0 ~V$ :

Vcommon = 10⁻⁵ \times 6 + 010⁻⁵ + 2 \times 10⁻⁵ = 6 \times 10⁻⁵3 \times 10⁻⁵ = 2 ~V

Calculating final charge on capacitor Q ( $C₂$ $C_2$):

Q₂ = C₂ Vcommon = (20 \times 10⁻⁶ ~F) \times 2 ~V = 40 \times 10⁻⁶ ~C = 4 \times 10⁻⁵ ~C

Comparing with $α \times 10⁻⁵ ~C α = 4$ .

Step 1: Final Conclusion

The value of $α$ $\alpha$ is 4.

Pattern Recognition

Charge distribution rule: Total initial charge $Qtotal = C₁ V₁ = 60$ . Final charge splits in proportion to capacitance ratio $C₂ / (C₁+C₂) = 2/3$ $C_2 / (C_1+C_2) = 2/3$. $Q₂ = (2/3) \times 60 = 40 = 4 \times 10⁻⁵~C$ .

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q44 jee_main_2026_28_january_morning Potential Energy of System of Charges

Two point charges of 1 nC and 2 nC are placed at the two corners of equilateral triangle of side 3 cm. The work done in bringing a charge of 3 nC from infinity to the third corner of the triangle is ____

.

14πε₀=9\times10⁹ N\cdot m²/C²

A. $2.7$ $2.7$
B. $5.4$ $5.4$
C. $3.3$ $3.3$
D. $27$ $27$

Solution

Related Formula

W = q₃ \cdot Vsystem

V = (kq₁)/(r₁) + (kq₂)/(r₂)

Core Logic

Work done is equal to the potential energy acquired by the 3 nC charge in the electric field of the other two charges.

Step 1: Formula Setup

W = ( (kq₁)/( ) + (kq₂)/( ) ) q₃

Step 2: Value Substitution

q₁ = 1 \times 10⁻⁹ ~C

q₂ = 2 \times 10⁻⁹ ~C

q₃ = 3 \times 10⁻⁹ ~C

= 3 \times 10⁻² ~m

W = 9 \times 10⁹3 \times 10⁻² ( 1 \times 10⁻⁹ + 2 \times 10⁻⁹ ) \times 3 \times 10⁻⁹

Step 3: Final Calculation

W = 3 \times 10¹¹ \times 3 \times 10⁻⁹ \times 3 \times 10⁻⁹

W = 27 \times 10⁻⁷ ~J = 2.7 \times 10⁻⁶ ~J = 2.7

Pattern Recognition

Factor out common terms ( $k/$ $k/\ell$) before substituting micro/nano orders to prevent simple arithmetic exponent errors.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q9 jee_main_2025_02_april_evening Electric Potential Difference

Two large plane parallel conducting plates are kept

10cm

apart as shown in figure. The potential difference between them is V. The potential difference between the points A and B (shown in the figure) is :

Parallel conducting plates with potential difference V showing points A and B — The diagram shows two parallel plates separated by 10 cm, with points A and B defining a right triangle of legs 3 cm and 4 cm.

A. $(1)/(4) ~V$ $\frac{1}{4} \mathrm{~V}$
B. $(2)/(5) ~V$ $\frac{2}{5} \mathrm{~V}$
C. $(3)/(4) ~V$ $\frac{3}{4} \mathrm{~V}$
D. $1 ~V$ $1 \mathrm{~V}$

Solution

Related Formula

Uniform Electric Field between parallel plates:

E = (V)/(d)

Potential difference between two points along the direction of the field:

Δ V = E \cdot Δ x

Core Logic

The separation between the plates is $d = 10 cm$ $d = 10 \ \mathrm{cm}$. The electric field $E$ $\vec{E}$ between them is uniform and runs perpendicularly between the plates (along the horizontal direction, $+x$ $+x$):

E = (V)/(10) V/cm

From the geometry shown in the figure:

$AC = 3 cm$ $AC = 3 \ \mathrm{cm}$ is perpendicular to $E$ $\vec{E}$ (vertical direction).
$CB = 4 cm$ $CB = 4 \ \mathrm{cm}$ is parallel to $E$ $\vec{E}$ (horizontal direction).

Step 1: Calculate potential difference between A and B

Since line $AC$ $AC$ is perpendicular to $E$ $\vec{E}$:

VA - VC = 0 VA = VC

Thus, the potential difference between $A$ $A$ and $B$ $B$ is completely due to the horizontal displacement $CB$ $CB$:

VAB = VA - VB = E \cdot (CB)

Substitute the values:

VAB = ((V)/(10)) \times 4 = (2)/(5) V

Thus, the potential difference between $A$ $A$ and $B$ $B$ is $(2)/(5)V$ $\frac{2}{5}V$.

Pattern Recognition

Sees: Uniform electric field, potential difference over diagonal paths. Trap: Projecting along the hypotenuse length ( $5 cm$ $5\text{ cm}$) directly without considering the electric field's physical direction. Shortcut: Electric field is purely horizontal. Hence, only horizontal displacement matters. The horizontal displacement is $4 cm$ $4\text{ cm}$ out of $10 cm$ $10\text{ cm}$ total plate gap, so the potential difference is $(4)/(10)V = (2)/(5)V$ $\frac{4}{10}V = \frac{2}{5}V$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q8 jee_main_2025_03_april_evening Combination of Capacitors and Charge Sharing

Using a battery, a

100~pF

capacitor is charged to

60~V

and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is

20~V

, its capacitance is: (in pF)

A. 600
B. 200
C. 400
D. 100

Solution

Related Formula

When a charged capacitor $C₁$ $C_1$ with voltage $V₁$ $V_1$ is connected in parallel to an uncharged capacitor $C₂$ $C_2$, the common final potential $Vc$ $V_c$ is determined by the conservation of charge:

Vc = (C₁ V₁ + C₂ V₂)/(C₁ + C₂)

Since $C₂$ $C_2$ is initially uncharged ( $V₂ = 0$ $V_2 = 0$):

Vc = (C₁ V₁)/(C₁ + C₂)

Core Logic

Given parameters:

First capacitor $C₁ = 100~pF$ $C_1 = 100\mathrm{~pF}$
Initial potential $V₁ = 60~V$ $V_1 = 60\mathrm{~V}$
Common final voltage $Vc = 20~V$ $V_c = 20\mathrm{~V}$

Step 1: Solve for $C₂$ $C_2$

Substitute the values into the common potential expression:

20 = (100 \times 60)/(100 + C₂)

20(100 + C₂) = 6000

100 + C₂ = (6000)/(20) = 300

C₂ = 300 - 100 = 200~pF

Pattern Recognition

Think of common potential as dilution. The potential drops to $1/3$ $1/3$ of its initial value ( $20~V/60~V$ $20\mathrm{~V}/60\mathrm{~V}$). This requires the total capacitance to triple ( $3 \times C₁$ $3 \times C_1$). Since they are in parallel, the added capacitor must be $2 \times C₁ = 200~pF$ $2 \times C_1 = 200\mathrm{~pF}$.

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

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Formula Setup

Step 2: Value Substitution

Step 3: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate potential difference between A and B

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for C₂$C_2$

Pattern Recognition

Chapter Mix

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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?

Step 1: Solve for $C₂$ $C_2$