1. Electric Charges and Fields
normal

Two charges ${q_1}$ and ${q_2}$ are placed $30\,\,cm$ apart, shown in the figure. A third charge ${q_3}$ is moved along the arc of a circle of radius $40\,cm$ from $C$ to $D$. The change in the potential energy of the system is $\frac{{{q_3}}}{{4\pi {\varepsilon _0}}}k$, where $k$ is

A$8\,{q_2}$
B$8\,{q_1}$
C$6\,{q_2}$
D$6\,{q_1}$

Solution

(a) Change in potential energy ($\Delta$$U$)$ = Uf -Ui$
$==>$ $\Delta U = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\left( {\frac{{{q_1}{q_3}}}{{0.4}} + \frac{{{q_2}{q_3}}}{{0.1}}} \right) – \left( {\frac{{{q_1}{q_3}}}{{0.4}} + \frac{{{q_2}{q_3}}}{{0.5}}} \right)} \right]$
$==>$ $\Delta U = \frac{1}{{4\pi {\varepsilon _0}}}[8{q_2}{q_3}] = \frac{{{q_3}}}{{4\pi {\varepsilon _0}}}(8{q_2})$
$k = 8q_2$
Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.