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Four metallic plates are used to form two identical parallel plate capacitors as shown is the figure. Initially all plates were unchanged. Two batteries are connected to two plates. What is magnitude of work done by electric field in moving electron from $x$ and $y$ ?

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3 eV

5 eV

3.5 eV

7 eV

answer is C.

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Detailed Solution

Charge on each capacitor is

$q = \frac{3 C}{2}$

Electric field between plates $= \frac{q}{A ε_{\circ}}$

$V_{x} + \frac{E d}{2} - 5 + \frac{E d}{2} = V_{y} V_{y} - V_{x} = E d - 5 = \frac{q}{ε_{0} A} d - 5 = \frac{q}{c} - 5 = \frac{\frac{3 C}{2}}{C} - 5 = \frac{3}{2} - 5 = - 3.5 V W_{e l} = q (V_{y} - V_{x}) = - (e) (- 3.5) = 3.5 e V$

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