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Question

(B) The work done $W$ to bring a charge $Q$ from infinity to the origin is given by $W = V \cdot Q$,where $V$ is the electric potential at the origin

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The work done $W$ to bring a charge $Q$ from infinity to the origin is given by $W = V \cdot Q$,where $V$ is the electric potential at the origin due to the four charges.The distances of the four charges from the origin $(0, 0)$ are:$r_1 = \sqrt{0^2 + 2^2} = 2$$r_2 = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$$r_3 = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$$r_4 = \sqrt{0^2 + (-2)^2} = 2$The potential $V$ at the origin is:$V = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q}{r_1} + \frac{Q}{r_2} + \frac{Q}{r_3} + \frac{Q}{r_4} \right)$$V = \frac{Q}{4\pi \varepsilon_0} \left( \frac{1}{2} + \frac{1}{2\sqrt{5}} + \frac{1}{2\sqrt{5}} + \frac{1}{2} \right)$$V = \frac{Q}{4\pi \varepsilon_0} \left( 1 + \frac{2}{2\sqrt{5}} \right) = \frac{Q}{4\pi \varepsilon_0} \left( 1 + \frac{1}{\sqrt{5}} \right)$Thus,the work done is $W = V \cdot Q = \frac{Q^2}{4\pi \varepsilon_0} \left( 1 + \frac{1}{\sqrt{5}} \right)$.

Four equal point charges $Q$ 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

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