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>>Electrostatic Potential and Capacitance
>>Parallel Plate Capacitor
>> $In an isolated parallel plate capacitor$

Question

In an isolated parallel plate capacitor of capacitance C the four surfaces have charges $Q_{1}, Q_{2}, Q_{3}$ and $Q_{4}$ as shown in the figure. The potential difference between the plates is :

112169

This question has multiple correct options

A

$\frac{Q _{1} + Q _{2}}{C}$

B

$∣ \frac{Q _{2}}{C} ∣$

C

$∣ \frac{Q _{3}}{C} ∣$

D

$\frac{1}{C} [(Q_{1} + Q_{2}) - (Q_{3} - Q_{4})]$

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Updated on : 2022-09-05

Solution

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Correct options are B) and C)

For parallel plate capacitor, $Q_{2}$ and $Q_{3}$ must be same and equal in magnitude.i.e, $∣ Q_{2} ∣ = ∣ Q_{3} ∣$

So, potential difference between the plates, $V = \frac{∣ Q _{2} ∣}{C} = \frac{∣ Q _{3} ∣}{C}$

Ans: $(B), (C)$

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