A two point charge 4q and -q are fixed on the x-axis at x=-\frac{d}{2} and x=\frac{d}{2}, respectively. If a third point charge 'q' is taken from the origin to x= d along the semicircle as shown in the figure, the energy of the charge will:
Option: 1 increased by \frac{3q^{2}}{4 \pi \epsilon _{0}d}
Option: 2 increase by \frac{2q^{2}}{3 \pi \epsilon _{0}d}
Option: 3 decreases by \frac{q^{2}}{4 \pi \epsilon _{0}d}
Option: 4 decrease by \frac{4q^{2}}{3 \pi \epsilon _{0}d}

Answers (1)

\begin{aligned} &\text { Initial and final potential energy are, }\\ &\mathrm{U}_{\mathrm{i}}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{4 \mathrm{q}^{2}}{\left(\frac{\mathrm{d}}{2}\right)}-\frac{\mathrm{q}^{2}}{\left(\frac{d}{2}\right)}\right]\\ &\mathrm{U}_{\mathrm{f}}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{4 \mathrm{q}^{2}}{\left(\frac{3 \mathrm{~d}}{2}\right)}-\frac{\mathrm{q}^{2}}{\left(\frac{d}{2}\right)}\right]\\ &\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=\Delta \mathrm{U}=(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4 q^2 }{(3 \mathrm{~d} / 2)})-(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4 q^2}{(d / 2)})\\ &=\frac{4 q^{2}}{4 \pi \varepsilon_{0}}\left(\frac{2}{3d}-\frac{2}{d}\right)\\ &=(-) \frac{4 q^{2}}{3 \pi \varepsilon_{0} \cdot d} \end{aligned}

 the energy of the charge will decrease by \frac{4q^{2}}{3 \pi \epsilon _{0}d}

 

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