First slide

Capacitance

Question

Four plates are located at a distance 'd' apart from one another. The extreme plates are interconnected. A potential difference of V is applied to the inner plates. The electric field E between '1' and '2' plates is :

Easy

A

$\frac{V}{4 d}$
B

$\frac{V}{2 d}$
C

$\frac{V}{d}$
D

$\frac{2 V}{d}$

Solution

The circuit can be redrawn as

Since, $V_{AB} = V$
So, $V_{OB} = \frac{V}{2}$
So, $E_{OB} = \frac{V}{2 d}$

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