A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A . If the rate of change of potential difference between the plates is $7 \times 10^{8} \mathrm{V} / \mathrm{s}$ then the integer value of the distance between the parallel plates is - $\left(\right.$ Take, $\left.\in_{0}=9 \times 10^{-12} \frac{\mathrm{~F}}{\mathrm{~m}}, \pi=\frac{22}{7}\right)$ $\_\_\_\_$ $\mu \mathrm{m}$.

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Capacitance

A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A . If the rate of change of potential difference between the plates is

7 \times 1 0^{8} V / s

then the integer value of the distance between the parallel plates is -

(

Take,

\in_{0} = 9 \times 1 0^{- 12} \frac{F}{m}, π = \frac{22}{7})

____

μ m

Answer: 1320

Solution

For a parallel plate capacitor of plate radius

r

and separation

d

C = \frac{ϵ _{0} π r ^{2}}{d}

With charging current

i

, the charge at time

t

Q = i t

, so

V = \frac{Q}{C} = \frac{i t d}{ϵ _{0} π r ^{2}}

Since

i

d

r

are constant during charging,

\frac{d V}{d t} = \frac{i d}{ϵ _{0} π r ^{2}} ⟹ d = \frac{ϵ _{0} π r ^{2}}{i} \cdot \frac{d V}{d t}

Substituting

ϵ_{0} = 9 \times 1 0^{- 12}

F/m,

r = 0.1

i = 0.15

\frac{d V}{d t} = 7 \times 1 0^{8}

V/s,

π = \frac{22}{7}

d = \frac{( 9 \times 1 0 ^{- 12} ) ( \frac{22}{7} ) ( 0.1 ) ^{2}}{0.15} \times (7 \times 1 0^{8}) = 1320 \times 1 0^{- 6} m = 1320 μ m

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Integer answer

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JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Capacitance

Solution

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