The capacitance of a parallel plate capacitor is C. If a dielectr
A. KC / 2(K + 1)
B. 2KC / K + 1
C. 5KC / 4K + 1
D. 4KC / 3K + 1
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
Concept:
The capacitance for a parallel plate capacitor with a dielectric constant ϵ is defined as:
\(C = \frac{{A\epsilon}}{d}\)
d = separation between the plates
ϵ = dielectric constant
A = Area of the plates
Also, for two capacitors in series, the net capacitance is given by:
\(\frac{1}{{{C_{eq}}\;}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}\)
\({C_{eq}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}\)
Application:
Given capacitance of the capacitor = C, i.e.
∴ \(C = \frac{{A\epsilon}}{d}\) …1)
Now, the dielectric slab of thickness one-fourth of the plate is inserted. This can be represented as:
This is nothing but two capacitors connected in series, as shown:
\({C_1} = \frac{{A\epsilon}}{{\left( {d - \frac{d}{4}} \right)}} = \frac{{4A\epsilon}}{{3d}}\;\;\)
Using equation (1), we can write:
\({C_1} = \frac{{4C}}{3}\)
The capacitance C2 of the dielectric will be:
\({C_2} = \frac{{Ak\epsilon}}{{d/4}} = \frac{{4AK\epsilon}}{d}\)
Using equation (1), we can write:
C2 = 4 KC
The equivalent capacitance will be:
\({C_{eq}} = \frac{{\frac{{4C}}{3}\; \times \;4KC}}{{\frac{4}{3}\;C\; + \;4KC}}\)
\({C_{eq}} = \frac{{16K{C^2}}}{{3\left( {\frac{{4C\; + \;12KC}}{3}} \right)}}\)
\({C_{eq}} = \frac{{16K{C^2}}}{{4C\; + \;12\;KC}}\)
\({C_{eq}} = \frac{{4KC}}{{1\; + \;3K}}\)
