Question 850: Electrostats & Capacitance

Question

Two point charges '$$q 1$$' and '$$q 2$$' are separated by a distance '$$d$$'. What is the increase in potential energy of the system when '$$q 2$$' is moved towards '$$q 1$$' by a distance '$$\mathrm{x}$$' ?
$$(x < d)(\frac{1}{4 \pi \varepsilon_0}=K$$, constant)

Accepted Answer

The potential energy between two charges is given as $$U=\frac{k q_1 q_2}{r}$$
Initial potential energy is $$U_{\mathrm{f}}=\frac{k q_1 q_2}{r}$$
When charge $$q_2$$ moves towards the $$q_1$$ the separation between the charges becomes $$\mathrm{d}-\mathrm{x}$$
The final potential energy is $$U_{\mathrm{f}}=\frac{\mathrm{kq}_1 \mathrm{q}_2}{(\mathrm{~d}-\mathrm{x})}$$
The increase in potential energy is
$$\begin{aligned}
& 	herefore \quad \Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{f}} \
& 	herefore \quad \Delta U=\frac{k q_1 q_2}{d}-\frac{k_1 q_2}{(d-x)} \
& 	herefore \quad \Delta U=\mathrm{kq}_1 \mathrm{q}_2\left(\frac{1}{\mathrm{~d}}-\frac{1}{\mathrm{~d}-\mathrm{x}}ight) \
& 	herefore \quad \Delta U=\frac{-\mathrm{kq}_1 \mathrm{q}_2 \mathrm{x}}{\mathrm{d}(\mathrm{d}-\mathrm{x})} \
\end{aligned}$$

Answer

$$-\frac{K q_1 q_2 x}{d(d-x)}$$

Answer

$$-\frac{K q_1 q_2}{d(d-x)}$$

Answer

$$\frac{\mathrm{Kq}_1 \mathrm{q}_{2 x}}{\left(\mathrm{~d}^2-\mathrm{x}^2\right)}$$

Answer

$$\frac{\mathrm{Kq}_1 \mathrm{q}_2 \mathrm{x}}{\left(\mathrm{d}^2-\mathrm{x}^2\right)}$$

Electrostats & Capacitance

Two point charges '$q 1' and 'q 2' are separated by a distance 'd'. What is the increase in potential energy of the system when 'q 2' is moved towards 'q 1' by a distance '\mathrm{x}' ? (x < d)(\frac{1}{4 \pi \varepsilon_0}=K$, constant)

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