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Physics Question on Capacitors and Capacitance

A parallel plate capacitor of value 1.77 μ\mu F is to be designed using a dielectric material (dielectric constant = 200), breakdown strength of 3×1063\times {{10}^{6}} V/m. In order to make such a capacitor which can withstand a potential difference of 20 V across the plates, the separation between the plates d and area A of the plates respectively are

A

6.6×10−6m,103m26.6\times {{10}^{-6}}m,{{10}^{3}}{{m}^{2}}

B

6.6×10−5m,104m26.6\times {{10}^{-5}}m,{{10}^{4}}{{m}^{2}}

C

6.6×10−4m, 105m26.6\times {{10}^{-4}}m,\text{ }{{10}^{5}}{{m}^{2}}

D

6.6×10−6m, 102m26.6\times {{10}^{-6}}m,\text{ }{{10}^{2}}{{m}^{2}}

Answer

6.6×10−6m,103m26.6\times {{10}^{-6}}m,{{10}^{3}}{{m}^{2}}

Explanation

Solution

We knows, electric field is given by
E=VdE=\frac{V}{d}
Emax⁡=Vdmin⁡{{E}_{\max }}=\frac{V}{{{d}_{\min }}}
dmin⁡=VEmax⁡{{d}_{\min }}=\frac{V}{{{E}_{\max }}}
Substitute V=20 VV=20\text{ }V and Emax⁡=3×106V{{E}_{\max }}=3\times {{10}^{6}}V
dmin⁡=203×106=6.6×10−6m{{d}_{\min }}=\frac{20}{3\times {{10}^{6}}}=6.6\times {{10}^{-6}}m
We knows, the capacitance is given by
C=kε0AdC=\frac{k{{\varepsilon }_{0}}A}{d}
Ad=Ckε0\frac{A}{d}=\frac{C}{k{{\varepsilon }_{0}}}
A=Ckε0A=\frac{C}{k{{\varepsilon }_{0}}}
Putting C=1.77 μF=1.77×10−6C,C=1.77\text{ }\mu F=1.77\times {{10}^{-6}}C,
d=6.6×10−6,k=200,d=6.6\times {{10}^{-6}},k=200,
ε0=8.85×10−12C2N−1m2{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{2}}
A=1.77×10−6×6.6×10−6200×8.85×10−12=103m2A=\frac{1.77\times {{10}^{-6}}\times 6.6\times {{10}^{-6}}}{200\times 8.85\times {{10}^{-12}}}={{10}^{3}}{{m}^{2}}