Two charges {q_1} and {q_2} are placed 30,cm | vedclass

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(A) The potential energy of a system of charges is given by $U = \frac{1}{4\pi \varepsilon_0} \sum \frac{q_i q_j}{r_{ij}}$.Initial potential energy at

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$8\,{q_2}$

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(A) The potential energy of a system of charges is given by $U = \frac{1}{4\pi \varepsilon_0} \sum \frac{q_i q_j}{r_{ij}}$.Initial potential energy at $C$ $(U_i)$: The distance of $q_3$ from $q_1$ is $0.4\,m$ and from $q_2$ is $0.5\,m$ (using Pythagoras theorem: $\sqrt{0.3^2 + 0.4^2} = 0.5\,m$).$U_i = \frac{1}{4\pi \varepsilon_0} \left( \frac{q_1 q_3}{0.4} + \frac{q_2 q_3}{0.5} + \frac{q_1 q_2}{0.3} \right)$.Final potential energy at $D$ $(U_f)$: The distance of $q_3$ from $q_1$ is $0.4\,m$ and from $q_2$ is $0.1\,m$.$U_f = \frac{1}{4\pi \varepsilon_0} \left( \frac{q_1 q_3}{0.4} + \frac{q_2 q_3}{0.1} + \frac{q_1 q_2}{0.3} \right)$.Change in potential energy $\Delta U = U_f - U_i = \frac{1}{4\pi \varepsilon_0} \left( \frac{q_2 q_3}{0.1} - \frac{q_2 q_3}{0.5} \right) = \frac{q_3}{4\pi \varepsilon_0} \left( 10 q_2 - 2 q_2 \right) = \frac{q_3}{4\pi \varepsilon_0} (8 q_2)$.Comparing this with the given expression $\frac{q_3}{4\pi \varepsilon_0} k$,we get $k = 8 q_2$.

Two charges ${q_1}$ and ${q_2}$ are placed $30\,cm$ apart,as 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:

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