Question
A parallel plate capacitor is made of two circular plate separated by a distance 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $$3 \times {10^4}V/m$$ the charge density of the positive plate will be close to:
A.
$$6 \times {10^{ - 7}}\,C/{m^2}$$
B.
$$3 \times {10^{ - 7}}\,C/{m^2}$$
C.
$$3 \times {10^4}\,C/{m^2}$$
D.
$$6 \times {10^4}\,C/{m^2}$$
Answer :
$$6 \times {10^{ - 7}}\,C/{m^2}$$
Solution :
Electric field in presence of dielectric between the two plates of a parallel plate capacitor is given by,
$$E = \frac{\sigma }{{K{\varepsilon _0}}}$$
Then, charge density
$$\eqalign{
& \sigma = K{\varepsilon _0}E \cr
& = 2.2 \times 8.85 \times {10^{ - 12}} \times 3 \times {10^4} \approx 6 \times {10^{ - 7}}C/{m^2} \cr} $$