JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Capacitance

A parallel plate capacitor whose capacitance C is 14 pF is charged by a battery to a potential difference

V = 12 V

between its plates. The charging battery is now disconnected and a porcelain plate with

k = 7

is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of ______ pJ. (Assume no friction)

Answer: 864

Solution

With the battery still connected the stored energy is

U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} (14) (12)^{2} = 1008 pJ

Once the battery is removed the charge on the plates is fixed. Inserting the slab raises the capacitance to

k C

, so with

U = \frac{Q ^{2}}{2 C}

U_{f} = \frac{U _{i}}{k} = \frac{1008}{7} = 144 pJ

The electrical energy given up is converted into kinetic energy of the slab, and with no friction this energy is conserved as the slab moves back and forth:

E = 1008 - 144 = 864 pJ

The constant mechanical energy of the oscillating plate is

864

pJ.

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Capacitance

Solution

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