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

A potential difference of

300 V

is applied between the plates of a plane capacitor spaced

1 cm

apart. A plane parallel glass plate with a thickness of

0.5 cm

and a plane parallel paraffin plate with a thickness of

0.5 cm

are placed in the space between the capacitor plates find (i) intensity of electric field in each layer (ii) the drop of potential in each layer (iii) the surface charge density of the charge on the plates. Given that

K_{paraffin} = 2, K_{glass} = 6

.

Accepted Answer

The half of the capacitor is filled with glass and half with paraffin. Let electric field inside the capacitor without dielectric be $1 cm$ , The electric field inside the dielectric is $\frac{E}{K}$ , The potential difference is $V = E \times d$ , Therefore the net potential difference across the capacitor $V = \frac{E_{0}}{K_{g}} d_{1} + \frac{E_{0}}{K_{p}} d_{2}$ , $d_{1} = d_{2} = 0.5 cm$ $\Rightarrow 300 = \frac{E_{0}}{6} \times 0.5 \times 10^{- 2} + \frac{E_{0}}{2} \times 0.5 \times 10^{- 2} \Rightarrow E_{0} = 9 \times 10^{4} V m^{- 1}$ , The intensity of electric field in glass layer $E_{g} = \frac{9 \times 10^{4}}{6} = 1.5 \times 10^{4} V m^{- 1}$ , Intensity of electric field in paraffin $E_{p} = \frac{9 \times 10^{4}}{2} = 4.5 \times 10^{4} V m^{- 1}$ ,

(ii). The potential drop across glass layer $V_{g} = 1.5 \times 10^{4} \times 0.5 \times 10^{- 2} V \Rightarrow V_{g} = 75 V$ . The potential drop across paraffin layer $V_{p} = 4.5 \times 10^{4} \times 0.5 \times 10^{- 2} V \Rightarrow V_{g} = 225 V$ (iii). For surface charge density $E = \frac{σ}{ε_{0}} \Rightarrow σ = E ε_{0}$

$\Rightarrow σ_{g} = 1.5 \times 10^{4} \times 8.85 \times 10^{- 12} \Rightarrow σ_{g} = 1.32 \times 10^{- 7} C m^{- 2} \Rightarrow σ_{p} = 4.5 \times 10^{4} \times 8.85 \times 10^{- 12} \Rightarrow σ_{g} = 3.98 \times 10^{- 7} C m^{- 2}$

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