- Home
- Standard 12
- Physics
2. Electric Potential and Capacitance
easy
A charge of $10\, e.s.u.$ is placed at a distance of $2\, cm$ from a charge of $40\, e.s.u.$ and $4\, cm$ from another charge of $20\, e.s.u.$ The potential energy of the charge $10\, e.s.u.$ is (in $ergs$)
A
$87.5$
B
$112.5$
C
$150$
D
$250$
Solution
(d) Energy $ = \frac{{10 \times 40}}{2} + \frac{{10 \times 20}}{4} = 250\,\,erg$
Standard 12
Physics
Similar Questions
normal