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 ABCD of side 10cm. What is the force on the charge of $1 μ c$ placed at the centre of the square.

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$12.72 \times 10^{9} N$

$15.91 \times 10^{9} N$

$2.545 \times 10^{9} N$

Zero

answer is A.

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Detailed Solution

Given data: A square ABCD with a side 10cm has four- point charges

$q_{A} = 2 μ C, q_{B} = - 5 μ C, q_{C} = 2 μ C and q_{D} = - 5 μ C$

Concept solution: Electric field and Elecytic field lines

Detailed solution:

Four charges are positioned at the corners of a square with a side of 10 cm, as shown in the figure shown. The square’s centre is O.

$sides = A B = B C = C D = A D = 10 cm$

$diagonals = A C = B D = 12 \sqrt{2} cm$

$A O = O C = D O = O B = 5 \sqrt{2} cm$

At point $O$ , a charge in the amount of $1 μ C$ is made.

In comparison to the force of repulsion between the charges located at corner C and centre O, the force of repulsion between the charges placed at corner A and centre O is similar in size but goes in the other direction. As a result, they will cancel one another.

As with the force of attraction between the charges put at corner D and centre O, the force of attraction between the charges placed at corner b and centre O is identical in strength but goes in the other direction. They will therefore cancel each other as well.

The net force produced by the four charges placed at the square’s corners on the 1C charge at centre 'O' is therefore Zero.