Name: Dielectrics in a Capacitor
Uploaded: 2019-07-19
Duration: 6 min 3 s
Description: Let us use our knowledge about dielectrics and capacitors to find out how the use of dielectrics can increase the capacitance of a system. Learn through this...

Question

The plates of a parallel plate capacitor are separated by a distance

d = 1 cm

. Two parallel sided dielectric slabs of thickness

0.7 cm

and

0.3 cm

fill the space between the plates. If the dielectric constants of the two slabs are

3

and

5

respectively and a potential difference of

440 V

is applied across the plates Find: (i) the electric field intensities in eahc of the slabs. (ii) the ratio of electric energies stored in the first to that in the second dielectric slab.

Accepted Answer

Some part of the capacitor is filled with dielectric $K = 3$ and other with dielectric $K = 5$ . Let electric field inside the capacitor without dielectric be $E_{0}$ , The electric filed inside the dielectric is $\frac{E}{K}$ , The potential difference is $3$ , Therefore the net potential difference across the capacitor $V = \frac{E_{0}}{K_{1}} d_{1} + \frac{E_{0}}{K_{2}} d_{2}$ , $d_{1} = 0.7 cm & d_{2} = 0.3 cm$ $\Rightarrow 440 = \frac{E_{0}}{3} \times 0.7 \times 10^{- 2} + \frac{E_{0}}{5} \times 0.3 \times 10^{- 2} \Rightarrow E_{0} = 1.5 \times 10^{5} V m^{- 1}$ , The intensity of electric field in first part $E_{1} = \frac{1.5 \times 10^{5}}{3} = 5 \times 10^{4} V m^{- 1}$ , Intensity of electric field in second part of the capacitor $E_{2} = \frac{1.5 \times 10^{5}}{5} = 3 \times 10^{4} V m^{- 1}$ ,

The potential drop across first part $V_{1} = 5 \times 10^{4} \times 0.7 \times 10^{- 2} V \Rightarrow V_{1} = 350 V$ . The potential drop across second part $V_{2} = 3 \times 10^{4} \times 0.3 \times 10^{- 2} V \Rightarrow V_{2} = 90 V$ .

(ii). The electric energy stored in capacitor $\frac{U_{1}}{U_{2}} = \frac{\frac{1}{2} C_{1} {V_{1}}^{2}}{\frac{1}{2} C_{2} {V_{2}}^{2}} \Rightarrow \frac{\frac{ε_{0} K_{1} A}{d_{1}} {V_{1}}^{2}}{\frac{ε_{0} K_{2} A}{d_{2}} {V_{2}}^{2}} \Rightarrow \frac{K_{1} {V_{1}}^{2} d_{2}}{K_{2} {V_{2}}^{2} d_{1}}$ ,

$\Rightarrow \frac{U_{1}}{U_{2}} = \frac{3 \times {(350)}^{2} \times 0.3}{5 \times {(90)}^{2} \times 0.7} \Rightarrow \frac{U_{1}}{U_{2}} = \frac{35}{9}$

Important Questions on Capacitance

Learn and Prepare for any exam you want !