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Question

# A parallel-plate capacitor is charged by a battery which is then disconnected. A dielectric slab is then inserted to fill the space between the plates. Explain the changes, if any, that occur in the values of (i) charge on the plates, (ii) electric field between the plates, (iii) p.d. between the plates, (iv) capacitance and (v) energy stored in the capacitor.

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Solution

## Dear student, When dielectric is placed in between the plates of capacitor and external electric field is applied then each molecule of the dielectric will get polarised it decreases the electric filed but increase the capacitance. Enet = Eexternal-Epolarised ${E}_{effective}=\frac{\sigma }{k{\epsilon }_{0}}=\frac{E}{k}\phantom{\rule{0ex}{0ex}}As,Thepotentialdifferenceacrossacapacitorisproportionalto\phantom{\rule{0ex}{0ex}}theelectricfieldbetweentheplates.\phantom{\rule{0ex}{0ex}}{V}_{effective}=\frac{V}{k}\phantom{\rule{0ex}{0ex}}Capaci\mathrm{tan}ce\phantom{\rule{0ex}{0ex}}{C}_{effective}=\frac{k{\epsilon }_{0}A}{d}\phantom{\rule{0ex}{0ex}}Energystored=\frac{1}{2}C{\left(∆V\right)}^{2}\phantom{\rule{0ex}{0ex}}AsCincreaseswithkeeping∆Vcons\mathrm{tan}t,Energyalsoincreasesby\epsilon .$ Regards

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