Question:
If three electric charges each of magnitude 20 µC are placed at any three corners of a square of side $\sqrt{2}\,m$, then the net electric field at the centre of the square (in $10^{5}NC^{-1}$) is}
If three electric charges each of magnitude 20 µC are placed at any three corners of a square of side $\sqrt{2}\,m$, then the net electric field at the centre of the square (in $10^{5}NC^{-1}$) is}
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Whenever identical charges occupy all corners of a symmetric figure, first check whether symmetry can simplify the calculation before resolving vectors.
Updated On: Jun 17, 2026
- 1.2
- 5.4
- 3.6
- 1.8
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The Correct Option is D
Solution and Explanation
Concept:
The electric field due to a point charge is
\[
E=\frac{kq}{r^2}
\]
where \(k=9\times10^9\,Nm^2C^{-2}\).
For four identical charges placed at the four corners of a square, the electric field at the centre becomes zero because of symmetry.
Therefore, the resultant field due to three charges is equal in magnitude to the field due to the missing fourth charge.
Step 1: Find the distance of the centre from a corner.
Given side of square, \[ a=\sqrt2\,m \] Distance of centre from a corner is half of the diagonal. \[ r=\frac{a\sqrt2}{2} \] \[ r=\frac{\sqrt2\times\sqrt2}{2} \] \[ r=1\,m \]
Step 2: Calculate electric field due to one charge.
Given, \[ q=20\mu C=20\times10^{-6}C \] Hence, \[ E=\frac{kq}{r^2} \] \[ E=\frac{9\times10^9\times20\times10^{-6}}{1^2} \] \[ E=1.8\times10^5\,NC^{-1} \]
Step 3: Apply symmetry argument.
Since the field due to four identical charges at the centre is zero, \[ \vec E_1+\vec E_2+\vec E_3+\vec E_4=0 \] Therefore, \[ \vec E_1+\vec E_2+\vec E_3=-\vec E_4 \] Thus the magnitude of the resultant field due to three charges is equal to the field due to one charge. \[ E_{net}=1.8\times10^5\,NC^{-1} \]
Step 4: Express in the required form.
\[ E_{net}=1.8\times10^5\,NC^{-1} \] Hence, \[ \boxed{1.8} \]
Step 1: Find the distance of the centre from a corner.
Given side of square, \[ a=\sqrt2\,m \] Distance of centre from a corner is half of the diagonal. \[ r=\frac{a\sqrt2}{2} \] \[ r=\frac{\sqrt2\times\sqrt2}{2} \] \[ r=1\,m \]
Step 2: Calculate electric field due to one charge.
Given, \[ q=20\mu C=20\times10^{-6}C \] Hence, \[ E=\frac{kq}{r^2} \] \[ E=\frac{9\times10^9\times20\times10^{-6}}{1^2} \] \[ E=1.8\times10^5\,NC^{-1} \]
Step 3: Apply symmetry argument.
Since the field due to four identical charges at the centre is zero, \[ \vec E_1+\vec E_2+\vec E_3+\vec E_4=0 \] Therefore, \[ \vec E_1+\vec E_2+\vec E_3=-\vec E_4 \] Thus the magnitude of the resultant field due to three charges is equal to the field due to one charge. \[ E_{net}=1.8\times10^5\,NC^{-1} \]
Step 4: Express in the required form.
\[ E_{net}=1.8\times10^5\,NC^{-1} \] Hence, \[ \boxed{1.8} \]
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