Question:
Charges \(q\), \(2q\), \(3q\) and \(4q\) are placed at the corners \(A\), \(B\), \(C\) and \(D\) respectively of a square as shown in the figure. The direction of electric field at the centre \(O\) of the square is along
Charges \(q\), \(2q\), \(3q\) and \(4q\) are placed at the corners \(A\), \(B\), \(C\) and \(D\) respectively of a square as shown in the figure. The direction of electric field at the centre \(O\) of the square is along
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When charges are placed at the corners of a square, first combine the electric fields due to opposite corners along each diagonal. Then add the two diagonal resultants using symmetry and vector addition.
Updated On: Jul 9, 2026
- AB
- CB
- BD
- AC \bigskip
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The Correct Option is B
Solution and Explanation
Concept:
The electric field at the centre due to a positive charge is directed away from the charge. Since all four charges are at the same distance from the centre, the magnitude of each electric field is proportional to the charge.
\[
E\propto q.
\]
Step 1: Resolve the fields along the diagonals. Charges at opposite corners lie on the same diagonal. Along diagonal \(AC\): \[ E_{AC}\propto (3q-q)=2q, \] directed from \(C\) towards \(A\). Along diagonal \(BD\): \[ E_{BD}\propto (4q-2q)=2q, \] directed from \(D\) towards \(B\).
Step 2: Add the two resultant fields. The two fields have equal magnitudes and act along the two diagonals of the square. Taking components, \[ \vec E_{AC} = (-\sqrt2\,E_0,\,-\sqrt2\,E_0), \] \[ \vec E_{BD} = (\sqrt2\,E_0,\,-\sqrt2\,E_0). \] Hence, \[ \vec E = \vec E_{AC}+\vec E_{BD} = (0,\,-2\sqrt2\,E_0). \] Thus the horizontal components cancel and the resultant field is vertically downward.
Step 3: Identify the direction. The downward direction through the centre is parallel to side \(CB\). Therefore, the electric field at the centre is along \[ CB. \] Final Answer: \[ \boxed{\text{Direction of electric field is along }CB} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Resolve the fields along the diagonals. Charges at opposite corners lie on the same diagonal. Along diagonal \(AC\): \[ E_{AC}\propto (3q-q)=2q, \] directed from \(C\) towards \(A\). Along diagonal \(BD\): \[ E_{BD}\propto (4q-2q)=2q, \] directed from \(D\) towards \(B\).
Step 2: Add the two resultant fields. The two fields have equal magnitudes and act along the two diagonals of the square. Taking components, \[ \vec E_{AC} = (-\sqrt2\,E_0,\,-\sqrt2\,E_0), \] \[ \vec E_{BD} = (\sqrt2\,E_0,\,-\sqrt2\,E_0). \] Hence, \[ \vec E = \vec E_{AC}+\vec E_{BD} = (0,\,-2\sqrt2\,E_0). \] Thus the horizontal components cancel and the resultant field is vertically downward.
Step 3: Identify the direction. The downward direction through the centre is parallel to side \(CB\). Therefore, the electric field at the centre is along \[ CB. \] Final Answer: \[ \boxed{\text{Direction of electric field is along }CB} \] \[ \boxed{\text{Answer = (B)}} \]
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