If four charges $q_1=+1 \times 10^{-8} \mathrm{C}, q_2=-2 \times 10^{-8} \mathrm{C}$, $q_3=+3 \times 10^{-8}…

If four charges $q_1=+1 \times 10^{-8} \mathrm{C}, q_2=-2 \times 10^{-8} \mathrm{C}$, $q_3=+3 \times 10^{-8} \mathrm{C}$ and $q_4=+2 \times 10^{-8} \mathrm{C}$ are kept at the four corners of a square of side 1 m , then the electric potential at the centre of the square is
  1. 300 V
  2. 200 V
  3. 510 V
  4. 410 V

Solution

$\begin{aligned} & \text { } \mathrm{q}_1=+1 \times 10^{-8} \mathrm{C}, \mathrm{q}_2=-2 \times 10^{-8} \mathrm{C} \\ & \mathrm{q}_3=+3 \times 10^{-8} \mathrm{C}, \mathrm{q}_4=+2 \times 10^{-8} \mathrm{C} \\ & \mathrm{a}=1 \mathrm{~m}\end{aligned}$

$\therefore \quad r=\frac{\sqrt{2} a}{2}=\frac{a}{\sqrt{2}}=\frac{1}{\sqrt{2}} \mathrm{~m}$
The electric potential at centre 0 is $\begin{aligned} & \mathrm{V}_0=\frac{\mathrm{k}}{\mathrm{r}}\left(\mathrm{q}_1+\mathrm{q}_2+\mathrm{q}_3+\mathrm{q}_4\right) \\ & \frac{9 \times 10^9}{\frac{1}{\sqrt{2}}}(1-2+3+2) \times 10^{-8} \\ & =509 \mathrm{~V} \approx 510 \mathrm{~V} \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

Practice more Electrostatics questions on Aicharya