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

• A. \(\dfrac{1}{2}\)
• B. \(\dfrac{1}{\sqrt{2}}\)
• C. \(\dfrac{1}{2\sqrt{2}}\)
• D. \(\dfrac{1}{4\sqrt{2}}\)

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Write the potential at the centre of a square with corner charges.
For a square of side $a$, distance from centre to each corner: $r = \frac{a}{\sqrt{2}}$. Total potential from four charges $q$: $V = \frac{4kq}{a/\sqrt{2}} = \frac{4kq\sqrt{2}}{a}$.
Step 2: Express $V_1$ for the first square ($a_1 = 1$ m).
\[ V_1 = \frac{4kq\sqrt{2}}{1} = 4kq\sqrt{2} \]
Step 3: Express $V_2$ for the second square ($a_2 = 2$ m, same charges).
\[ V_2 = \frac{4kq\sqrt{2}}{2} = 2kq\sqrt{2} \]
Step 4: Compute the ratio.
\[ \frac{V_2}{V_1} = \frac{2kq\sqrt{2}}{4kq\sqrt{2}} = \frac{1}{2} \]
Step 5: Interpret: potential scales inversely with side length.
Doubling the side doubles each corner-to-centre distance, halving each charge's contribution and the total potential.
Step 6: State the final answer.
\[ \boxed{\frac{V_2}{V_1} = \frac{1}{2}} \]