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 \(5\,\text{cm}\).
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 \(5\,\text{cm}\).
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In asymmetric charge configurations, resolve components instead of assuming cancellation.
Updated On: Jun 16, 2026
- at \(45^\circ\) to OA upward
- at \(135^\circ\) to OA towards BD
- no direction, because \(E = 0\)
- None of the above
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The Correct Option is A
Solution and Explanation
Concept:
Electric field at centre due to each charge is along the line joining the charge and centre.
Step 1: Charges
\[ A(+q),\; B(-2q),\; C(-q),\; D(+2q) \]
Step 2: Directions of fields
• Positive charge → field away from charge
• Negative charge → field towards charge
Step 3: Vector addition
Resolve all four contributions: - Fields from opposite corners do not completely cancel due to unequal magnitudes - Net resultant points in upward diagonal direction
Step 4: Final direction
Resultant electric field is along a direction making \(45^\circ\) with OA upward. \[ \therefore \text{Answer is (A)} \]
Step 1: Charges
\[ A(+q),\; B(-2q),\; C(-q),\; D(+2q) \]
Step 2: Directions of fields
• Positive charge → field away from charge
• Negative charge → field towards charge
Step 3: Vector addition
Resolve all four contributions: - Fields from opposite corners do not completely cancel due to unequal magnitudes - Net resultant points in upward diagonal direction
Step 4: Final direction
Resultant electric field is along a direction making \(45^\circ\) with OA upward. \[ \therefore \text{Answer is (A)} \]
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