Let $\mathrm{V}_1$ be the potential at the center of the square of side $1 \mathrm{~m}$ when the charges at…

Let $\mathrm{V}_1$ be the potential at the center of the square of side $1 \mathrm{~m}$ when the charges at the 4 corners are $2 \mathrm{C}$ each. If the same charges are placed at the corners of a square of side $2 \mathrm{~m}$, then the potential at the centre of this square is $\mathrm{V}_2$. The value of $\frac{V_2}{V_1}$ is
  1. $\frac{1}{2}$
  2. $\frac{1}{\sqrt{2}}$
  3. $\frac{1}{2 \sqrt{2}}$
  4. $\frac{1}{4 \sqrt{2}}$

Solution


$\left(\mathrm{V}_{\text {centre }}\right)_1=\frac{\mathrm{K} \times 2}{\frac{1}{\sqrt{2}}} \times 4=\mathrm{V}_1$
$ \left(\mathrm{V}_{\text {centre }}\right)_2=\frac{\mathrm{K} \times 2}{\frac{2}{\sqrt{2}}} \times 4=\mathrm{V}_2 $ So, $\frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{1}{2}$

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

Practice more Electrostatics questions on Aicharya