1.

The potential energy between plates of a charged parallel plate capacitor is UQ. If a dielectric plate of dielectric constant er is placed between The gap of plates, then the new potential energy will be :(a) \(\frac{U_{0}}{\varepsilon_{r}}\)(b) U0εr2(c) \(\frac{U_{0}}{\varepsilon_{r}^{2}}\)(d) U0

Answer»

(a) \(\frac{U_{0}}{\varepsilon_{r}}\)
Energy decreases to εr times
U = \(\frac{U_{0}}{\varepsilon_{r}}\)



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