Two charges of 10 μ C are placed at the corners A and B of an equilateral triangle of side length 0.5 m in air. The electric potential at C is

Question

Two charges of 10 μ C are placed at the corners A and B of an equilateral triangle of side length 0.5 m in air. The electric potential at C is (A) 36 × 104 V (B) 5 × 106 V (C) 7 × 109 V (D) 107 V

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/81a952ca1e555485c039cff982be4c50-.png" /> According to the principle of superposition V net = V AC + V BC V net =( Kq / r )+( Kq / r ) V net =(2 Kq / r ) V text net =(2 × 9 × 109 × 10 × 10-6/0.5 m ) =36 × 104 V

Q. Two charges of $10 μ C$ are placed at the corners $A$ and $B$ of an equilateral triangle of side length $0.5 m$ in air. The electric potential at $C$ is

$36 \times 1 0^{4} V$

$5 \times 1 0^{6} V$

$7 \times 1 0^{9} V$

$1 0^{7} V$

According to the principle of superposition
$V_{n e t} = V_{A C} + V_{BC}$
$V_{n e t} = \frac{K q}{r} + \frac{K q}{r}$
$V_{n e t} = \frac{2 K q}{r}$
$V_{net} = \frac{2 \times 9 \times 1 0 ^{9} \times 10 \times 1 0 ^{- 6}}{0.5 m}$
$= 36 \times 1 0^{4} V$

Q. Two charges of 10μC are placed at the corners A and B of an equilateral triangle of side length 0.5m in air. The electric potential at C is

36×104V

5×106V

7×109V

107V