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A parallel plate condenser is immersed in an oil of dielectric constant 2 . The field between the plates is
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Decreased proportional to $\frac{1}{2}$
Let before immersed in oil the capacitance is \(\mathrm{C}\), potential is \(\mathrm{V}\). and field, \(\mathrm{E}=\mathrm{V} / \mathrm{d}\) where \(\mathrm{d}=\) separation of plates.
After immersed in oil the capacitance becomes \(\mathrm{C}^{\prime}=\mathrm{kC}=2 \mathrm{C}\) and potential becomes \(\mathrm{V}^{\prime}=\mathrm{V} / 2\) because \(\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}\).
Now filed, \(E^{\prime}=V^{\prime} / d=\frac{1}{2}(V / d)=\frac{E}{2}\)
After immersed in oil the capacitance becomes \(\mathrm{C}^{\prime}=\mathrm{kC}=2 \mathrm{C}\) and potential becomes \(\mathrm{V}^{\prime}=\mathrm{V} / 2\) because \(\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}\).
Now filed, \(E^{\prime}=V^{\prime} / d=\frac{1}{2}(V / d)=\frac{E}{2}\)
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