- Home
- Standard 12
- Physics
2. Electric Potential and Capacitance
easy
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 $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,\frac{{N{\rm{ - }}{m^2}}}{{{C^2}}}} \right]$
A
$9 \times {10^4}\,V$
B
$18 \times {10^4}\,V$
C
$36 \times {10^4}\,V$
D
$36 \times {10^{ - 4}}\,V$
Solution
(c) Potential at C $ = \left( {9 \times {{10}^9} \times \frac{{4 \times {{10}^{ – 6}}}}{{0.2}}} \right) \times 2= 36 \times 10^4 \,V$
Standard 12
Physics
