$A$ proton is about $1840$ times heavier than an electron. When it is accelerated by a potential difference of $1\, kV$,its kinetic energy will be......$keV$.

Two particles $A$ and $B$ having the same mass have charges $+q$ and $+4q$ respectively. When they are allowed to fall from rest through the same electric potential difference,the ratio of their speeds $v_A$ to $v_B$ will be:

What is the electric potential at a distance of $9 \text{ cm}$ from a charge of $3 \text{ nC}$ (in $\text{ V}$)?

If a charged spherical conductor of radius $10 \,cm$ has potential $V$ at a point distant $5 \,cm$ from its centre, then the potential at a point distant $15 \,cm$ from the centre will be

$A$ charge $+q$ is fixed at each of the points $x = x_0, x = 3x_0, x = 5x_0, \dots$ up to $\infty$ on the $X$-axis and a charge $-q$ is fixed at each of the points $x = 2x_0, x = 4x_0, x = 6x_0, \dots$ up to $\infty$. Here $x_0$ is a positive constant. Taking the potential at a point due to a charge $Q$ at a distance $r$ from it to be $\frac{Q}{4\pi\varepsilon_0 r}$,find the potential at the origin due to the above system of charges.

$A$ cube of side $b$ has a charge $q$ at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.