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Four Charges +Q, −Q, +Q And −Q Are to Be Arranged Respectively at the Four Corners of a Square Abcd of Side 'A'. - Physics

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Question

Four charges +q, −q, +q and −q are to be arranged respectively at the four corners of a square ABCD of side 'a'.
(a) Find the work required to put together this arrangement.
(b) A charge q0 is brought to the centre of the square, the four charges being held fixed. How much extra work is needed to do this ?

Solution

Here,
AB = BC = CD = DA = a
And,

\[AC = BD = a\sqrt{2}\]

Now,

\[\text{ Total work done} \left( W \right) = \text { Potential energy of the system of four charges } \]

\[W = \frac{\left( - q \right)\left( + q \right)}{4 \pi\epsilon_0 a} + \frac{\left( + q \right)\left( - q \right)}{4 \pi\epsilon_0 a} + \frac{\left( - q \right)\left( + q \right)}{4 \pi\epsilon_0 a} + \frac{\left( + q \right)\left( - q \right)}{4 \pi\epsilon_0 a} + \frac{1}{4 \pi\epsilon_0}\frac{\left( \left( - q \right)\left( - q \right) \right)}{a\sqrt{2}} + \frac{1}{4 \pi\epsilon_0}\frac{\left( \left( + q \right)\left( + q \right) \right)}{a\sqrt{2}}\]

\[ = \frac{- 4 q^2}{4 \pi\epsilon_0 a} + \frac{2 q^2}{4 \pi\epsilon_0 a\sqrt{2}} = \frac{- q^2}{4 \pi\epsilon_0 a}\left( 4 - \sqrt{2} \right)\]

\[\left( b \right)\]

\[\text { Extra work needed,} W_P = q_0 \times V_P \]

\[\text { Here,} \]

\[V_P \text{= Potential at point P }\]

\[\text { Now,} \]

\[\text { Potential at point P } = \frac{q}{4\pi \epsilon_0 \frac{a\sqrt{2}}{2}} + \frac{\left( - q \right)}{4\pi \epsilon_0 \frac{a\sqrt{2}}{2}} + \frac{q}{4\pi \epsilon_0 \frac{a\sqrt{2}}{2}} + \frac{\left( - q \right)}{4\pi \epsilon_0 \frac{a\sqrt{2}}{2}} = 0\]

\[ \Rightarrow W = 0\]

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2014-2015 (March) Foreign Set 2

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