A total charge Q is given to two identical capacitors connected in parallel. Initial separation of plate in each capacitor is d0. Then suddenly the first plate of first capacitor & second plate of second capacitor simultaneously starts moving towards left with speed v as shown in figure. Which of following statement(s) are correct at time t.

Engineering

Physics

Parallel Plate Capacitor

Electric Flux and Gauss Law and its Applications

Basic Circuit Theory and Kirchoffs Law for DC Circuit

Question

A total charge Q is given to two identical capacitors connected in parallel. Initial separation of plate in each capacitor is d₀. Then suddenly the first plate of first capacitor & second plate of second capacitor simultaneously starts moving towards left with speed v as shown in figure. Which of following statement(s) are correct at time t.

$I_{t} = \frac{Qv}{2 d_{0}}$

$q_{upper plate} = \frac{Q (d_{0} - vt)}{2 d_{0}}$

$I_{t} = \frac{Qv}{d_{0}}$

$q_{lower plate} = \frac{Q (d_{0} + vt)}{d_{0}}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Let q, & q₂ be the instantaneous charges on capacitors :

$\frac{q_{1}}{C_{1}} = \frac{q_{2}}{C_{1}} & q_{1} + q_{2} = Q$

$C_{1} = \frac{ε_{0} A}{d_{0} + vt}, C_{2} = \frac{ε_{0} A}{d_{0} - vt}$

$\frac{q_{1}}{q_{2}} = \frac{C_{1}}{C_{2}} = \frac{d_{0} - vt}{d_{0} + vt}$

$q_{1} = Q (\frac{d_{0} - vt}{2 d_{0}}) : q_{2} = Q (\frac{d_{0} + vt}{2 d_{0}})$

$I = \frac{{dq}_{2}}{dt} = - \frac{{dq}_{1}}{dt} = \frac{Qv}{2 d_{0}}$