JEE MAIN - Physics (2014 (Offline) - No. 15)
A parallel plate capacitor is made of two circular plates 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:
$$6 \times {10^{ - 7}}\,\,C/{m^2}$$
$$3 \times {10^{ - 7}}\,\,C/{m^2}$$
$$3 \times {10^4}\,\,C/{m^2}$$
$$6 \times {10^4}\,\,C/{m^2}$$
Explanation
Electric field in presence of dielectric between the two plates of a parallel plate capacitor is given by,
$$E = {\sigma \over {K{\varepsilon _0}}}$$
Then, charge density
$$\sigma = K{\varepsilon _0}E$$
$$ = 2.2 \times 8.85 \times {10^{ - 12}} \times 3 \times {10^4} \approx 6 \times {10^{ - 7}}\,\,C/{m^2}$$
$$E = {\sigma \over {K{\varepsilon _0}}}$$
Then, charge density
$$\sigma = K{\varepsilon _0}E$$
$$ = 2.2 \times 8.85 \times {10^{ - 12}} \times 3 \times {10^4} \approx 6 \times {10^{ - 7}}\,\,C/{m^2}$$
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