Question

Two charges of $4\mu C$ each are placed at the corners A and B of an equilateral triangle of side length 0.2 m in air. The electric potential at C is $\left\lbrack \frac{1}{4\pi\varepsilon_{0}} = 9 \times 10^{9}\frac{N - m^{2}}{C^{2}} \right\rbrack$

• A. $9 \times 10^{4}$V
• B. $18 \times 10^{4}$V
• C. $36 \times 10^{4}$V
• D. $36 \times 10^{- 4}$

Answer 1

Answer: (C)

$36 \times 10^{4}$V

Answer 2

Potential at C $= \left( 9 \times 10^{9} \times \frac{4 \times 10^{- 6}}{0.2} \right) \times 2$

= 36 × 10⁴ V