Four equal point charges Q are placed in the xy-plane at positions (0,2), (4,2), (4,-2) and (0,-2) (in metres). What is the work done in bringing a fifth charge Q from infinity to the origin?

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EEJ MAIN Physics QUESTION #6994

Question 1

Four equal point charges $Q$ are placed in the $xy$-plane at positions $(0,2)$, $(4,2)$, $(4,-2)$ and $(0,-2)$ (in metres). What is the work done in bringing a fifth charge $Q$ from infinity to the origin?

$\dfrac{Q^2}{4\pi\varepsilon_0}\left(1 + \dfrac{1}{\sqrt{3}}\right)$
$\dfrac{Q^2}{4\pi\varepsilon_0}\left(1 + \dfrac{1}{\sqrt{5}}\right)$✔️
$\dfrac{Q^2}{2\sqrt{2}\,\pi\varepsilon_0}$
$\dfrac{Q^2}{4\pi\varepsilon_0}$

Correct Answer Explanation

Work done = $Q \times V_{origin}$, where $V_{origin}$ is the potential at origin due to the four charges.

Distances from origin:

$(0,2)$: $r_1 = 2$
$(0,-2)$: $r_2 = 2$
$(4,2)$: $r_3 = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$
$(4,-2)$: $r_4 = \sqrt{20} = 2\sqrt{5}$

$V = \frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2\sqrt{5}}+\frac{1}{2\sqrt{5}}\right) = \frac{Q}{4\pi\varepsilon_0}\left(1 + \frac{1}{\sqrt{5}}\right)$

Work $= QV = \dfrac{Q^2}{4\pi\varepsilon_0}\left(1+\dfrac{1}{\sqrt{5}}\right)$

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