If $3$ charges are placed at the vertices of equilateral triangle of charge ‘$q$’ each. What is the net potential energy, if the side of equilateral triangle is $l\, cm$
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q^2}}}{l}$
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{2{q^2}}}{l}$
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{3{q^2}}}{l}$
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{4{q^2}}}{l}$
In the figure shown the electric potential energy of the system is: ( $q$ is at the centre of the conducting neutral spherical shell of inner radius $a$ and outer radius $b$ )
On moving a charge of $20$ coulombs by $2 \;cm , 2 \;J$ of work is done, then the potential difference between the points is (in $volt$)
An elementary particle of mass $m$ and charge $ + e$ is projected with velocity $v$ at a much more massive particle of charge $Ze,$ where $Z > 0.$What is the closest possible approach of the incident particle
Charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively which are a distance $2\,L$ apart, $C$ is the midpoint between $A$ and $B.$ The work done in moving a charge $+Q$ along the semicircle $CRD$ is
An $\alpha $-particle is accelerated through a potential difference of $200\,V$. The increase in its kinetic energy is.......$eV$