Question

Three charges each of \(+4\mu \text{C}\) are placed at the corners A, B, C of a square ABCD of side \(1\) m. The electric field at the centre of the square is

• A. \(7.2\times 10^{4}\) N/C towards point B
• B. \(7.2\times 10^{4}\) N/C towards point D
• C. \(3.6\times 10^{4}\) N/C towards point A
• D. \(3.6\times 10^{4}\) N/C towards point C

Hint:

Charges at opposite corners cancel at the centre, leaving only the field of the third charge.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The field at the centre is the vector sum of the fields of the three equal charges at A, B and C. A and C are opposite corners.

Step 2: Cancellation
A and C are at the same distance from the centre in opposite directions, so their fields cancel.

Step 3: Field of B
Distance from corner to centre \(r=\dfrac{\sqrt{2}}{2}\) m, so \(r^{2}=0.5\) m\(^{2}\).
\(E=\dfrac{kq}{r^{2}}=\dfrac{9\times 10^{9}\times 4\times 10^{-6}}{0.5}=7.2\times 10^{4}\) N/C

Step 4: Direction
The field of a positive charge points away from it, so from B through the centre towards D. This is option (B).

Final Answer:
The field is \(7.2\times 10^{4}\) N/C directed towards point D. \[ \boxed{\text{(B) }7.2\times 10^{4}\ \text{N/C towards point D}} \]