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Q.

Four point charges Q_1,Q₂ , Q₃ and Q₄ are placed at the four corners A, B, C and D of a rectangle having sides 40 cm and 30 cm as shown in figure. Potential at the centre O is found to be 20 volt. If the charge Q₄ at corner D is removed, potential at O is found to be 2 volt. Then the value of Q₄ is

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a

0.5 nC

b

2.5 nC

c

3.75 nC

d

6 nC

answer is A.

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Detailed Solution

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$\begin{array}{l} Length AC = BD = \sqrt{30^{2} + 40^{2}} cm = 50 cm . \\ Now, ∴ V = \frac{{KQ}_{1}}{AO} + \frac{{KQ}_{2}}{BO} + \frac{{KQ}_{3}}{CO} + \frac{{KQ}_{4}}{DO} \\ and V'= \frac{{KQ}_{1}}{AO} + \frac{{KQ}_{2}}{BO} + \frac{{KQ}_{3}}{CO} \\ Subtracting, V - V' = \frac{{KQ}_{4}}{DO} \Rightarrow 20 - 2 = 9 \times 10^{9} \times \frac{Q_{4}}{0 .25} \Rightarrow Q_{4} = 0 .5 n C \end{array}$

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