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Four electric charges +q, +q, −q and −q are placed at the corners of a square of side 2L (see figure). The electric potential at point A, midway between the two charges +q and +q, is
A
Zero
B
$\frac{1}{4\pi\varepsilon_0}\frac{2q}{L}(1 + \sqrt{5})$
C
$\frac{1}{4\pi\varepsilon_0}\frac{2q}{L}\left(1 + \frac{1}{\sqrt{5}}\right)$
D
$\frac{1}{4\pi\varepsilon_0}\frac{2q}{L}\left(1 - \frac{1}{\sqrt{5}}\right)$
Detailed Solution
A is the midpoint of the side joining the two +q charges, so each +q is at a distance L from A.
Each −q charge is at the far corners: its distance from A is $\sqrt{(2L)^2 + L^2} = \sqrt{5}L$
Potential at A due to the two +q charges: $V_+ = 2\times\frac{1}{4\pi\varepsilon_0}\frac{q}{L} = \frac{1}{4\pi\varepsilon_0}\frac{2q}{L}$
Potential at A due to the two −q charges: $V_- = -2\times\frac{1}{4\pi\varepsilon_0}\frac{q}{\sqrt{5}L} = -\frac{1}{4\pi\varepsilon_0}\frac{2q}{\sqrt{5}L}$
$V_A = V_+ + V_- = \frac{1}{4\pi\varepsilon_0}\frac{2q}{L}\left(1 - \frac{1}{\sqrt{5}}\right)$
Each −q charge is at the far corners: its distance from A is $\sqrt{(2L)^2 + L^2} = \sqrt{5}L$
Potential at A due to the two +q charges: $V_+ = 2\times\frac{1}{4\pi\varepsilon_0}\frac{q}{L} = \frac{1}{4\pi\varepsilon_0}\frac{2q}{L}$
Potential at A due to the two −q charges: $V_- = -2\times\frac{1}{4\pi\varepsilon_0}\frac{q}{\sqrt{5}L} = -\frac{1}{4\pi\varepsilon_0}\frac{2q}{\sqrt{5}L}$
$V_A = V_+ + V_- = \frac{1}{4\pi\varepsilon_0}\frac{2q}{L}\left(1 - \frac{1}{\sqrt{5}}\right)$