On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved, and a dielectric medium of K = 10 is introduced between the plates, without disconnecting the d.c. source. Explain, using suitable expressions, how the (i) capacitance, (ii) electric field.

Solution

A parallel plate capacitor us charged to a potential V. After some time, spacing between the plates is halved and dielectric medium of K=10 is introduced without disconnecting the d.c source.

When the d.c source remains connected, potential across the plates remain same.

i) Initial capacitance, C= $\frac{ε_{\circ} A}{d}$
When the spacing between the plates is halved d=d/2
A dielectric slab of K=10 is inserted.
New capacitance, C'= K $\frac{ε_{0} A}{\frac{d}{2}}$ = $10 \frac{2 ε_{\circ} A}{d}$ =20 C
New capacitance becomes 20times that of the initial capacitance.

ii) Electric field is given by $E = \frac{V}{d}$
When spacing is halved.
New electric field becomes E'= $\frac{2 V}{d} = 2 E$

Therefore, electric field becomes twice that of the initial field.

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Electrostatic Potential And Capacitance

On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved, and a dielectric medium of K = 10 is introduced between the plates, without disconnecting the d.c. source. Explain, using suitable expressions, how the (i) capacitance, (ii) electric field.

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