JEE Main 2025 Jan 29 Shift 2, Physics Q22: Capacitance For Geometrical Figures With Dielectric

JEE Main2025Jan 29, Shift 2Physics

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 – $(Take, ε_{0} = 9 \times 1 0^{- 12} \frac{F}{m}, π = \frac{22}{7})$ _________ $μ$ m.

Your answer

Solution

Capacitance $C = \frac{ε _{0} π r ^{2}}{d}$ , and at constant charging current $I$ , $V = \frac{Q}{C} = \frac{I t}{C}$ , so

$\frac{d V}{d t} = \frac{I}{C} = \frac{I d}{ε _{0} π r ^{2}} \Rightarrow d = \frac{ε _{0} π r ^{2}}{I} \cdot \frac{d V}{d t}$ .

Substitute the given values: $r = 0.1$ m, $I = 0.15$ A, $d V / d t = 7 \times 1 0^{8}$ V/s, $ε_{0} = 9 \times 1 0^{- 12}$ F/m, $π = 22/7$ .

$d = \frac{( 9 \times 1 0 ^{- 12} ) \cdot ( 22/7 ) \cdot ( 0.1 ) ^{2}}{0.15} \cdot (7 \times 1 0^{8})$

$= \frac{9 \cdot 22 \cdot 0.01}{0.15} \times 1 0^{- 4} m = 13.2 \times 1 0^{- 4} m = 1320 μ m$ .

Concept behind this question

Capacitance For Geometrical Figures With DielectricNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 29 Shift 2, Physics Q22: Capacitance For Geometrical Figures With Dielectric

Concept behind this question

More previous year questions on Capacitance For Geometrical Figures With Dielectric

Practice more Physics