Particles $A$ and $B$ have charges $+q$ and $+4q$ respectively. Both have the same mass $m$. If they are released from rest and accelerated through the same potential difference $V$,the ratio of their velocities $v_A/v_B$ is:

Along the $X$-axis,three charges $\frac{q}{2}, -q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2a$ respectively. The resultant electric potential at $x=a+r$ (if $a << r$) is: ($\varepsilon_0$ is the permittivity of free space)

$A$ point charge $-q$ is carried from a point $A$ to another point $B$ on the axis of a charged ring of radius $r$ carrying a charge $+q$. If the point $A$ is at a distance $\frac{4}{3} r$ from the centre of the ring and the point $B$ is $\frac{3}{4} r$ from the centre but on the opposite side, what is the net work that needs to be done for this?

Charge is uniformly distributed on the surface of a hollow hemisphere. Let $O$ and $A$ be two points on the base of the hemisphere,where $O$ is the center of the base and $A$ is a point between the center and the rim. Let $V_O$ and $V_A$ be the electric potentials at $O$ and $A$ respectively. Then,

$A$ 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 $ergs$)

Four point charges $-Q, -q, 2q$ and $2Q$ are placed,one at each corner of a square. The relation between $Q$ and $q$ for which the potential at the center of the square is zero is: