Question

What would be the kinetic energy k2q of charge 2q at a very large distance from the other charges? express your answer in terms of q, d, and appropriate constants?

Answer 1

Answer:

$K_{2q}=\frac{7.76kq^2}{d}$

Explanation:

At the corner of the square, the potential energy of interaction of other charges with the charge 2q  is given by $U_{2q}$

So

$U_{2q,i}=k\frac{(2q)(q)}{d}+k\frac{(2q)(5q)}{d}+k\frac{(2q)(-3q)}{\sqrt{2}d}=\frac{7.76 kq^2}{d}$

Also, since $K_{2q,i}=0$

The initial energy of the system is given by;

$E_i=U_{2q ,i}+K_{2q,i}=\frac{7.76kq^2}{d}+0=\frac{7.76kq^2}{d}$

Since $U_{2q,f}=0$

, the final energy of the system is obtained by

$E_f=U_{2q ,f}+K_{2q,f}=0+K_{2q,f}$

From the law of conservation of energy, $E_i=E_f$

Therefore, $K_{2q}=\frac{7.76kq^2}{d}$