Question

What is the direction of the electric field at the centre \(O\) of the square in the figure shown below? Given that, \( q = 10\,\text{nC} \) and the side of the square is \(5\,\text{cm}\).

• A. at \(45^\circ\) to OA upward
• B. at \(135^\circ\) to OA towards BD
• C. no direction, because \(E = 0\)
• D. None of the above

Hint:

In asymmetric charge configurations, resolve components instead of assuming cancellation.

Answer 1

The Correct Option is A

The problem requires determining the direction of the electric field at the center \(O\) of a square. The charges at the corners of the square are \(+q\), \(-2q\), \(-q\), and \(+2q\) at points \(A\), \(B\), \(C\), and \(D\) respectively.

To find the resultant electric field at \(O\), consider the contributions from each charge:

1. E_A: Electric field due to charge \(+q\) at \(A\) acts away from \(A\) towards \(O\).
2. E_B: Electric field due to charge \(-2q\) at \(B\) acts towards \(B\).
3. E_C: Electric field due to charge \(-q\) at \(C\) acts towards \(C\).
4. E_D: Electric field due to charge \(+2q\) at \(D\) acts away from \(D\).

Now we calculate the net electric field at \(O\):

• The electric field vectors from \(A\) to \(O\) and \(C\) to \(O\) add constructively along the diagonal \(AC\).
• The electric field vectors from \(B\) to \(O\) and \(D\) to \(O\) also add along the diagonal \(BD\).

The charges and distances are arranged symmetrically such that the diagonals \(AC\) and \(BD\) are at \(45^\circ\) to \(OA\). The combined contributions from these vectors result in a net electric field along the diagonal \(AC\) at \(45^\circ\) to \(OA\) upward.

Therefore, the correct answer is:

at \(45^\circ\) to OA upward