Question-35 The electric charges +2q, +2q, -2q and -2q are placed at the corners of square of side 2L as shown in figure. The electric potential at point ‘A’, midway between the two charges +2q and +2q is (\varepsilon_0 = permittivity of free space)
(A) \frac{1}{4\pi\varepsilon_0}\left(\frac{2q}{L}\right)\left[1+\frac{1}{\sqrt{5}}\right]
(B) \frac{q}{\pi\varepsilon_0 L}\left[1-\frac{1}{\sqrt{5}}\right]
(C) \frac{q}{\pi\varepsilon_0 L}\left[1+\frac{1}{\sqrt{5}}\right]
(D) \frac{1}{4\pi\varepsilon_0}\left(\frac{2q}{L}\right)\left[1-\frac{1}{\sqrt{5}}\right]
Answer : (B)
Explanation:
AC² = (2L)^2 + L^2 = 4L^2 + L^2 = 5L^2
∴ AC = AD = \sqrt{5}L
AB = AE = L
Potential at point A due to the four charges
V = \frac{1}{4\pi \varepsilon_0}\left(\frac{2q}{L} + \frac{2q}{L} - \frac{2q}{\sqrt{5}L} - \frac{2q}{\sqrt{5}L}\right)
= \frac{1}{4\pi \varepsilon_0}\left(\frac{4q}{L} - \frac{4q}{\sqrt{5}L}\right)
= \frac{1}{4\pi \varepsilon_0}\cdot \frac{4q}{L}\left(1 - \frac{1}{\sqrt{5}}\right)
= \frac{q}{\pi \varepsilon_0 L}\left(1 - \frac{1}{\sqrt{5}}\right)