Question

Eight equal point charges each of charge $q$ and mass $m$ is placed at eight corners of a cube of side $a$ .

$(i)$ Find out potential energy of charge system

$(ii)$ Find out work done by an external agent against electrostatic forces and by electrostatic forces to increase all sides of the cube from $a$ to $2 a$ .

$(iii)$ If all the charges are released at rest, then find out their speed when they are at the corners of the cube of the side $2 a$ .

$(iv)$ If keeping all other charges fixed, the charge of the corner $A$ is released then find out its speed when it is at an infinite distance?

$(v)$ If all charges are released simultaneously from rest then find out their speed when they are at a very large distance from each other.

Accepted Answer

$(a)$ Potential energy of two charges $U = \frac{k q_{1} q_{2}}{r}$

In the cube one face has four charges at its corner, they have four sets at equal distance and two sets of arrangements diagonally.

Therefore, one face has six configurations.

The potential energy of the charges at one face.

$U_{face} = 4 \frac{k q^{2}}{a} + 2 \frac{k q q}{a \sqrt{2}}$

$\Rightarrow U_{face} = \frac{k q^{2}}{a} (4 + \sqrt{2})$ .

The cube has six faces

The total electric potential energy of the cube.

$U = 6 \times U_{face}$

$U = 6 \times \frac{k q^{2}}{a} (4 + \sqrt{2})$

$\Rightarrow U = \frac{6 k q^{2}}{a} (4 + \sqrt{2})$ .

$(b)$ Potential energy when sides are $2 a$

$\Rightarrow U^{'} = \frac{6 k q^{2}}{2 a} (4 + \sqrt{2})$

$\Rightarrow U^{'} = \frac{3 k q^{2}}{a} (4 + \sqrt{2})$

Work done by electrostatic force $= - ∆ U$

$∆ U = U^{'} - U$

$∆ U = \frac{3 k q^{2}}{a} (4 + \sqrt{2}) - \frac{6 k q^{2}}{a} (4 + \sqrt{2})$

$\Rightarrow ∆ U = - \frac{3 k q^{2}}{a} (4 + \sqrt{2})$

$W = \frac{3 k q^{2}}{a} (4 + \sqrt{2})$ .

Work done by external agent $= W_{ext} = ∆ U = - \frac{3 k q^{2}}{a} (4 + \sqrt{2})$ .

$(c)$ Work done by electric force when side of cube becomes $2 a$ from $a$ will be equal to increase in kinetic energy of all the charges.

$W = \frac{3 k q^{2}}{a} (4 + \sqrt{2}) = 8 \times \frac{1}{2} m v^{2}$

$4 m v^{2} = \frac{3 k q^{2}}{a} (4 + \sqrt{2})$

$\Rightarrow v = \frac{q}{2} \sqrt{\frac{3 k}{m a} (4 + \sqrt{2})}$ .

$(d)$ For the potential energy of the charge with other charges, there will be three pairs with the ends of edges, three with face diagonal and one with opposite corner diagonal.

$U = 3 \frac{k q q}{a} + 3 \frac{k q q}{\sqrt{2} a} + \frac{k q q}{\sqrt{3} a}$

$U = \frac{3 k q^{2}}{a} (1 + \frac{1}{\sqrt{2}} + \frac{1}{3 \sqrt{3}})$ .

Therefore, loss in electric potential energy of the charge $=$ gain in kinetic energy of the charge.

$\frac{3 k q^{2}}{a} (1 + \frac{1}{\sqrt{2}} + \frac{1}{3 \sqrt{3}}) = \frac{1}{2} m v^{2}$

$\Rightarrow v = q \sqrt{\frac{6 k}{m a} (1 + \frac{1}{\sqrt{2}} + \frac{1}{3 \sqrt{3}})}$ .

$(e)$ Loss in electric potential energy of all the charges $=$ gain in kinetic energy of all the charges

$\frac{6 k q^{2}}{a} (4 + \sqrt{2}) = 8 \times \frac{1}{2} m v^{2}$

$4 m v^{2} = \frac{6 k q^{2}}{a} (4 + \sqrt{2})$

$\Rightarrow v = \sqrt{\frac{3 k q}{2 m a} (4 + \sqrt{2})}$ .

Important Questions on Electrostatics

Learn and Prepare for any exam you want !