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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