JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Capacitance

A simple pendulum of mass '

m

', length '

ℓ

' and charge '

+ q

' suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be:

(A) $tan^{- 1} [\frac{q}{m g} \times \frac{C _{2} ( V _{2} - V _{1} )}{( C _{1} + C _{2} ) ( d - t )}]$
(B) $tan^{- 1} [\frac{q}{m g} \times \frac{C _{1} ( V _{1} + V _{2} )}{( C _{1} + C _{2} ) ( d - t )}]$ ✓ Correct
(C) $tan^{- 1} [\frac{q}{m g} \times \frac{C _{1} ( V _{2} - V _{1} )}{( C _{1} + C _{2} ) ( d - t )}]$
(D) $tan^{- 1} [\frac{q}{m g} \times \frac{C _{2} ( V _{1} + V _{2} )}{( C _{1} + C _{2} ) ( d - t )}]$

Answer: (B)

Solution

The plates are held at

+ V_{2}

and

- V_{1}

, so the potential difference across them is

V_{1} + V_{2}

. The dielectric slab of thickness

t

and the air gap of thickness

d - t

behave as two capacitors in series, of capacitances

C_{1}

and

C_{2}

respectively, carrying the same charge

Q = \frac{C _{1} C _{2}}{C _{1} + C _{2}} (V_{1} + V_{2})

The bob hangs in the air gap, where

C_{2} = \frac{ε _{0} A}{d - t}

, so the field there is

E = \frac{Q}{ε _{0} A} = \frac{Q}{C _{2} ( d - t )} = \frac{C _{1} ( V _{1} + V _{2} )}{( C _{1} + C _{2} ) ( d - t )}

In equilibrium the string tension balances gravity and the electric force,

T cos θ = m g

and

T sin θ = q E

, so

tan θ = \frac{q E}{m g}

and the deflection is

θ = tan^{- 1} [\frac{q}{m g} \times \frac{C _{1} ( V _{1} + V _{2} )}{( C _{1} + C _{2} ) ( d - t )}]

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Capacitance

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