First slide

Electrostatic potential

Question

Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, -2) and (0, -2). The work required to put a fifth charge Q at the origin of the coordinate system will be :

Easy

A

$\frac{Q^{2}}{4 π \in_{0}} (1 + \frac{1}{\sqrt{5}})$
B

$\frac{Q^{2}}{4 π \in_{0}} (1 + \frac{1}{\sqrt{3}})$
C

$\frac{Q^{2}}{2 \sqrt{2} π \in_{0}}$
D

$\frac{Q^{2}}{4 π \in_{0}}$

Solution

$W = Q V = Q (V_{1} + V_{2} + V_{3} + V_{4}) = Q [\frac{1}{4 π \in_{0}} \frac{Q}{r_{1}} + \frac{1}{4 π \in_{0}} \frac{Q}{r_{2}} + \frac{1}{4 π \in_{0}} \frac{Q}{r_{3}} + \frac{1}{4 π \in_{0}} \frac{Q}{r_{4}}] = \frac{1}{4 π \in_{0}} Q^{2} [\frac{1}{2} + \frac{1}{2} + \frac{1}{\sqrt{2^{2} + 4^{2}}} + \frac{1}{\sqrt{2^{2} + 4^{2}}}] = \frac{Q^{2}}{4 π \in_{0}} [1 + \frac{2}{\sqrt{20}}] = \frac{Q^{2}}{4 π \in_{0}} [1 + \frac{2}{2 \sqrt{5}}] = = \frac{Q^{2}}{4 π \in_{0}} [1 + \frac{1}{\sqrt{5}}]$

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