JEE PYQ: Electrostatics Question 57
Question 57 - 2019 (10 Jan Shift 2)
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:
(1) $\frac{Q^2}{4\pi\varepsilon_0}\left(1 + \frac{1}{\sqrt{3}}\right)$
(2) $\frac{Q^2}{4\pi\varepsilon_0}\left(1 + \frac{1}{\sqrt{5}}\right)$
(3) $\frac{Q^2}{2\sqrt{2}\pi\varepsilon_0}$
(4) $\frac{Q^2}{4\pi\varepsilon_0}$
Show Answer
Answer: (2)
Solution
Potential at origin due to four charges: distances are 2, $\sqrt{20}$, $\sqrt{20}$, 2. $V = KQ\left(\frac{1}{2} + \frac{1}{2} + \frac{1}{\sqrt{20}} + \frac{1}{\sqrt{20}}\right) = KQ\left(1 + \frac{1}{\sqrt{5}}\right)$. Work $= QV = \frac{Q^2}{4\pi\varepsilon_0}\left(1 + \frac{1}{\sqrt{5}}\right)$.
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