Answer:
[tex]K_{2q}=\frac{7.76kq^2}{d}[/tex]
Explanation:
At the corner of the square, the potential energy of interaction of other charges with the charge 2q is given by [tex]U_{2q}[/tex]
So
[tex]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}[/tex]
Also, since [tex]K_{2q,i}=0[/tex]
The initial energy of the system is given by;
[tex]E_i=U_{2q ,i}+K_{2q,i}=\frac{7.76kq^2}{d}+0=\frac{7.76kq^2}{d}[/tex]
Since [tex]U_{2q,f}=0[/tex]
, the final energy of the system is obtained by
[tex]E_f=U_{2q ,f}+K_{2q,f}=0+K_{2q,f}[/tex]
From the law of conservation of energy, [tex]E_i=E_f[/tex]
Therefore, [tex]K_{2q}=\frac{7.76kq^2}{d}[/tex]
