Two charges q_1 and q_2 are separated by a di | vedclass

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(A) The potential at point $C$ due to charges $q_1$ and $q_2$ is:$V_C = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{AC} + \frac{q_2}{BC} \right)$Give

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$8 q_2$

Vedclass · Answer

(A) The potential at point $C$ due to charges $q_1$ and $q_2$ is:$V_C = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{AC} + \frac{q_2}{BC} \right)$Given $AC = 40\ cm = 0.4\ m$. Since $AB = 30\ cm$ and $AC = 40\ cm$,by Pythagoras theorem,$BC = \sqrt{0.3^2 + 0.4^2} = 0.5\ m$.So,$V_C = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.5} \right)$.The potential at point $D$ due to charges $q_1$ and $q_2$ is:$V_D = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{AD} + \frac{q_2}{BD} \right)$Given $AD = 40\ cm = 0.4\ m$ (radius of the circular path) and $BD = AD - AB = 40\ cm - 30\ cm = 10\ cm = 0.1\ m$.So,$V_D = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.1} \right)$.The change in potential energy $\Delta U$ is given by:$\Delta U = q_3 (V_D - V_C) = q_3 \left[ \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.1} \right) - \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.5} \right) \right]$$\Delta U = \frac{q_3}{4 \pi \epsilon_0} \left( \frac{q_2}{0.1} - \frac{q_2}{0.5} \right) = \frac{q_3}{4 \pi \epsilon_0} (10 q_2 - 2 q_2) = \frac{8 q_2 q_3}{4 \pi \epsilon_0}$.Comparing this with $\frac{q_3 K}{4 \pi \epsilon_0}$,we get $K = 8 q_2$.

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