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Question

Two charges $$ 10 \mu C $$ and $$ 5 \mu C $$ are at $$ I m $$ apart. In order to make this distance $$ 0.5 \mathrm{m}, $$ how much work would be done?


Solution

$$\begin{array}{l}\mathrm{q}_{1}=10 \mu \mathrm{C}=10 \times 10^{-6} \mathrm{C} \\\mathrm{q}_{2}=5 \mu \mathrm{C}=5 \times 10^{-6} \mathrm{C} \\\dfrac{1}{4 \pi \varepsilon_{0}}=k=9 \times 10^{9} \dfrac{\mathrm{N} \cdot \mathrm{m}^{2}}{\mathrm{C}^{2}} \\r_{1}=1 \mathrm{m} \\r_{2}=0.5 \mathrm{m}\end{array}$$
First situation $$ U_{1}=\dfrac{k q_{1} q_{2}}{r_{1}} $$
Second situation $$ U_{2}=\dfrac{k q_{1} q_{2}}{r_{2}} $$
Work done $$ W=U_{2}-U_{1} $$
$$\begin{array}{l}W=\dfrac{k q_{1} q_{2}}{r_{2}}-\dfrac{k q_{1} q_{2}}{r_{1}} \\W=k q_{1} q_{2}\left[\dfrac{1}{r_{2}}-\dfrac{1}{r_{1}}\right] \\W=9 \times 10^{9} \times 10 \times 10^{-6} \times 5 \times 10^{-6}\left[\dfrac{1}{0.5}-\dfrac{1}{1}\right]\end{array}$$
$$ W=45 \times 10^{-2}\left[\dfrac{10}{5}-\dfrac{1}{1}\right] $$
$$ W=45 \times 10^{-2}[2-1] $$
$$ W=45 \times 10^{-2} \mathrm{J} $$

Physics

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