Question

A parallel plat capacitor is made of two circular plates separated by a distance of $5mm$ 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. $3\times {10}^{4}C/m^2$
• B. $6\times {10}^{4}C/m^2$
• C. $6\times {10}^{-7}C/m^2$
• D. $3\times {10}^{-7}C/m^2$

Answer 1

The correct option is C $6\times {10}^{-7}C/m^2$

Answer is C.
By formula of electric field between the plates of a capacitor $E=\frac{\sigma }{K{\epsilon }_0}$
$\sigma =EK{\epsilon }_0=3\times {10}^4\times 2.2\times 8.85\times {10}^{-12}$
$=6.6\times 8.85\times {10}^{-8}$
$=5.841\times {10}^{-7}$
$\cong 6\times {10}^{-7}C/m^2$