Question

Two identical particles each of mass m and having charges – q and +q are revolving in a circle of radius r under the influence of electric attraction. Kinetic energy of each particle is $\bigg \lgroup k=\frac{1}{4\pi\epsilon_{0}}\bigg \rgroup$
(a) $\frac{kq^{2}}{4r}$
(b) $\frac{kq^{2}}{2r}$
(c) $\frac{kq^{2}}{8r}$
(d) $\frac{kq^{2}}{r}$

Answer 1

Hey

Kinetic energy of each particle is $\bigg \lgroup k=\frac{1}{4\pi\epsilon_{0}}\bigg \rgroup$

(a) $\frac{kq^{2}}{4r}$

(b) $\frac{kq^{2}}{2r}$

(c) $\frac{kq^{2}}{8r}$

(d) $\frac{kq^{2}}{r}$

A correct

Answer 2

Two identical particles each of mass m and having charges – q and +q are revolving in a circle of radius r under the influence of electric attraction. Kinetic energy of each particle is $\bigg \lgroup k=\frac{1}{4\pi\epsilon_{0}}\bigg \rgroup$

(a) $</strong><strong>{</strong><strong>\</strong><strong>b</strong><strong>o</strong><strong>l</strong><strong>d</strong><strong>\frac{kq^{2}}{4r}</strong><strong>}</strong><strong>$

(b) $\frac{kq^{2}}{2r}$

(c) $\frac{kq^{2}}{8r}$

(d) $\frac{kq^{2}}{r}$