Four charges $\mathrm{q}_{1}=2 \times 10^{-8} \mathrm{C}, \mathrm{q}_{2}=-2 \times 10^{-8} \mathrm{C},…

Four charges $\mathrm{q}_{1}=2 \times 10^{-8} \mathrm{C}, \mathrm{q}_{2}=-2 \times 10^{-8} \mathrm{C}, \mathrm{q}_{3} =-3 \times 10^{-8} \mathrm{C}$, and $\mathrm{q}_{4}=6 \times 10^{-8} \mathrm{C}$ are placed at four corners of a square of side $\sqrt{2} \mathrm{~m}$. What is the potential at the centre of the square?
  1. $270 \mathrm{~V}$
  2. $300 \mathrm{~V}$
  3. Zero
  4. $100 \mathrm{~V}$

Solution

Potential at the centre $\mathrm{O}, \mathrm{V}=\mathrm{k} \frac{\mathrm{q}}{\mathrm{r}}$
$\mathrm{V}=\mathrm{k}\left[\frac{2 \times 10^{-8}}{1}+\frac{-2 \times 10^{-8}}{1}+\frac{-3 \times 10^{-8}}{1}+\frac{6 \times 10^{-8}}{1}\right]$
$\mathrm{V}=\mathrm{k} \times 3 \times 10^{-8}=9 \times 10^{9} \times 3 \times 10^{-8} \mathrm{volt}$
$=27 \times 10=270 \mathrm{~V}$

Asked in: JEE Mains - Electrostatics - Test 4

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