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Q 1.6: Four point charges $q_{A}= 2 \mu C$ , $q_{B}= -5 \mu C$ , $q_{C}= 2 \mu C$ , and $q_{D}= -5 \mu C$ are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

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Charges at (A, C) and (B, D) are pairwise diametrically opposite and also equal.

Therefore, their force on a point charge at the centre of the square will be equal but opposite in directions.

Now, AC = BD = $\sqrt{2}\times(10\times10^{-2} m) = \sqrt{2}\times0.1 m$

$\therefore$ AO = BO = CO = DO = r = Half of diagonal = $\sqrt{2}\times0.05 m$

Force on point charge at centre due to charges at A and C = $F_{A} = -F_{C} = \frac{k(2\mu C)(1\mu C)}{(r)^2}$

Similarly, Force on point charge at centre due to charges at B and D = $F_{B} = -F_{D} = \frac{k(-5\mu C)(1\mu C)}{(r)^2}$

$\therefore$ Net force on point charge = $\\F_{A} + F_{B} + F_{C} + F_{D} \\$

$= -F_{B} +F_{B} - F_{D} + F_{D} = 0$ .

Hence, the charge at the centre experiences no force.

Posted by

HARSH KANKARIA

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