Four point charges $q_{A} = 2 μ C, q_{B} = - 5 μ C, q_{C} = 2 μ C$ , and $q_{D} = - 5 μ C$ are located at the corners of a square $A B C D$ of side $10 c m$ . What is the force on a charge of $1 μ C$ placed at the centre of the square?

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Solution

The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.
Where,
(Sides) $A B = B C = C D = A D = 10 c m$
(Diagonals) $A C = B D = 10 \sqrt{2} c m$
$A O = O C = D O = O B = 5 \sqrt{2} c m$
A charge of amount $1 μ C$ is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on $1 μ C$ charge at centre O is zero.

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