Question:
Let \(V_1\) be the potential at the center of the square of side \(1\text{ m}\) when the charges at the \(4\) corners are \(2\text{ C}\) each. If the same charges are placed at the corners of a square of side \(2\text{ m}\), then the potential at the center of this square is \(V_2\). The value of \(\frac{V_2}{V_1}\) is
Let \(V_1\) be the potential at the center of the square of side \(1\text{ m}\) when the charges at the \(4\) corners are \(2\text{ C}\) each. If the same charges are placed at the corners of a square of side \(2\text{ m}\), then the potential at the center of this square is \(V_2\). The value of \(\frac{V_2}{V_1}\) is
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Electric potential is a scalar quantity. For equal charges at the corners of a square, add potentials directly, and remember that the center-to-corner distance is proportional to the side length.
Updated On: Jun 25, 2026
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{\sqrt{2}}\)
- \(\dfrac{1}{2\sqrt{2}}\)
- \(\dfrac{1}{4\sqrt{2}}\)
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The Correct Option is A
Solution and Explanation
Step 1: Formula for potential due to point charges.
Electric potential due to a point charge is \[ V=\frac{kq}{r} \] At the center of a square, all four corner charges are at the same distance from the center.
So, total potential is \[ V=4\frac{kq}{r} \]
Step 2: Find distance from center to corner.
For a square of side \(a\), diagonal is \[ a\sqrt{2} \] Distance from center to a corner is half the diagonal: \[ r=\frac{a\sqrt{2}}{2} \]
Step 3: Compare potentials for side \(1\text{ m}\) and \(2\text{ m}\).
Since \[ V=4\frac{kq}{r}, \] and \(k,q\) are same in both cases, we have \[ V\propto \frac{1}{r} \] Also, \[ r\propto a \] Therefore, \[ V\propto \frac{1}{a} \] For first square: \[ a_1=1\text{ m} \] For second square: \[ a_2=2\text{ m} \] Thus, \[ \frac{V_2}{V_1} = \frac{a_1}{a_2} \] \[ = \frac{1}{2} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{1}{2}} \]
Electric potential due to a point charge is \[ V=\frac{kq}{r} \] At the center of a square, all four corner charges are at the same distance from the center.
So, total potential is \[ V=4\frac{kq}{r} \]
Step 2: Find distance from center to corner.
For a square of side \(a\), diagonal is \[ a\sqrt{2} \] Distance from center to a corner is half the diagonal: \[ r=\frac{a\sqrt{2}}{2} \]
Step 3: Compare potentials for side \(1\text{ m}\) and \(2\text{ m}\).
Since \[ V=4\frac{kq}{r}, \] and \(k,q\) are same in both cases, we have \[ V\propto \frac{1}{r} \] Also, \[ r\propto a \] Therefore, \[ V\propto \frac{1}{a} \] For first square: \[ a_1=1\text{ m} \] For second square: \[ a_2=2\text{ m} \] Thus, \[ \frac{V_2}{V_1} = \frac{a_1}{a_2} \] \[ = \frac{1}{2} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{1}{2}} \]
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