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Electrostatic Potential and Capacitance · Question 2

JEE Advanced 2026 Paper 2NumericalHard📈 Topic in 17% papers+4
1

The difference in the capacitance (in F) between the metal plates at t=0 and that at t=500 s is (8-n) 0, where 0 is the permittivity of free space. The value of n is:

Answer

1.90 to 2.10

Solution

Key idea: each chamber (area 1 m2, gap 2 m) is a liquid layer h in series with air (2-h): C= 0Ah/ r+(2-h); the outside region is unchanged and cancels in the difference.

t=0: left = 02/15=7.5 0, right = 02=0.5 0; total 8 0.

t=500 s (h1=1.25, h2=0.75): left = 01.25/15+0.75=1.2 0, right = 00.75/15+1.25= 01.3=0.769 0.

C=8 0-1.969 0=(8-n) 0 n 1.97.

Answer: 1.97 (accepted range 1.90 to 2.10)

JEE Advanced 2026 Paper 2

Keep practising — Electrostatic Potential and Capacitance

2

A positive point charge of 10-8 C is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm. The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved by a distance of 10 cm further away from the center of the sphere along the radial direction. Taking 14 0=9 109 Nm2/C2 (where 0 is the permittivity of free space), which of the following statements is/are correct:

  1. Before the grounding, the electrostatic potential of the sphere is 450 V.
  2. Charge flowing from the sphere to the ground because of grounding is 5 10-9 C.
  3. After the grounding is removed, the charge on the sphere is -5 10-9 C.
  4. The final electrostatic potential of the sphere is 300 V.
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3

Four identical thin, square metal sheets, S1, S2, S3 and S4, each of side a are kept parallel to each other with equal distance d ( a) between them, as shown in the figure. Let C0= 0 a2/d, where 0 is the permittivity of free space. [figure] Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

  1. P → 3; Q → 2; R → 4; S → 5
  2. P → 2; Q → 3; R → 2; S → 1
  3. P → 3; Q → 2; R → 4; S → 1
  4. P → 3; Q → 2; R → 2; S → 5
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4

An infinitely long thin wire, having a uniform charge density per unit length of 5 nC/m, is passing through a spherical shell of radius 1 m, as shown in the figure. A 10 nC charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points P and R, in Volt, is . [Given: In SI units 14 0 = 9 109, 2 = 0.7. Ignore the area pierced by the wire.] [figure]

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