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

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(A) The initial potential energy $U_i$ of the system when $q_3$ is at $C$ is given by the sum of potential energies of all pairs:$U_i = \frac{1}{4\pi\

Vedclass · Accepted Answer

$8q_2$

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(A) The initial potential energy $U_i$ of the system when $q_3$ is at $C$ is given by the sum of potential energies of all pairs:$U_i = \frac{1}{4\pi\epsilon_0} \left[ \frac{q_1 q_3}{0.4} + \frac{q_1 q_2}{0.3} + \frac{q_2 q_3}{0.5} \right]$The final potential energy $U_f$ of the system when $q_3$ is at $D$ is:$U_f = \frac{1}{4\pi\epsilon_0} \left[ \frac{q_1 q_3}{0.4} + \frac{q_1 q_2}{0.3} + \frac{q_2 q_3}{0.1} \right]$The change in potential energy is $\Delta U = U_f - U_i$:$\Delta U = \frac{1}{4\pi\epsilon_0} \left[ \left( \frac{q_1 q_3}{0.4} + \frac{q_1 q_2}{0.3} + \frac{q_2 q_3}{0.1} \right) - \left( \frac{q_1 q_3}{0.4} + \frac{q_1 q_2}{0.3} + \frac{q_2 q_3}{0.5} \right) \right]$$\Delta U = \frac{q_3}{4\pi\epsilon_0} \left[ q_2 \left( \frac{1}{0.1} - \frac{1}{0.5} \right) \right]$$\Delta U = \frac{q_3}{4\pi\epsilon_0} \left[ q_2 (10 - 2) \right] = \frac{q_3}{4\pi\epsilon_0} (8q_2)$Comparing this with the given expression $\frac{q_3}{4\pi\epsilon_0}k$,we get $k = 8q_2$.

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 $\frac{q_3}{4\pi \epsilon_0}k$,where $k$ is:

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