Question 132: Electrostats & Capacitance

Question

A parallel plate capacitor was made with two rectangular plates, each with a length of $l=3 \mathrm{~cm}$ and breath of $\mathrm{b}=1 \mathrm{~cm}$. The distance between the plates is $3 \mu \mathrm{~m}$. Out of the following, which are the ways to increase the capacitance by a factor of 10 ?
A. $l=30 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=1 \mu \mathrm{~m}$
B. $l=3 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=30 \mu \mathrm{~m}$
C. $l=6 \mathrm{~cm}, \mathrm{~b}=5 \mathrm{~cm}, \mathrm{~d}=3 \mu \mathrm{~m}$
D. $l=1 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=10 \mu \mathrm{~m}$
E. $l=5 \mathrm{~cm}, \mathrm{~b}=2 \mathrm{~cm}, \mathrm{~d}=1 \mu \mathrm{~m}$
Choose the correct answer from the options given below:

Accepted Answer

The capacitance of a parallel plate capacitor is given by
$$ C = \frac{\epsilon_0 A}{d}, $$
where:
$$\epsilon_0$$ is the permittivity of free space,
$$A$$ is the area of a plate, and
$$d$$ is the separation between the plates.
A change in the capacitorÃ¢ÂÂs dimensions will change its capacitance by a factor of
$$ \frac{C_{	ext{new}}}{C_{	ext{initial}}} = \frac{A_{	ext{new}}}{A_{	ext{initial}}} \cdot \frac{d_{	ext{initial}}}{d_{	ext{new}}}. $$
The initial dimensions are:
Length: $$l = 3\,	ext{cm} = 0.03\,	ext{m},$$
Breadth: $$b = 1\,	ext{cm} = 0.01\,	ext{m},$$
Hence, the original area is
$$ A_{	ext{initial}} = 0.03 	imes 0.01 = 3 	imes 10^{-4}\,	ext{m}^2, $$
Plate separation: $$d_{	ext{initial}} = 3\,\mu	ext{m} = 3 	imes 10^{-6}\,	ext{m}.$$
For each modification, we compare the new factor:
Case C:
New dimensions:
$$l = 6\,	ext{cm} = 0.06\,	ext{m}, \quad b = 5\,	ext{cm} = 0.05\,	ext{m}.$$
New area:
$$      A_{	ext{new}} = 0.06 	imes 0.05 = 3 	imes 10^{-3}\,	ext{m}^2.      $$
Ratio of areas:
$$      \frac{A_{	ext{new}}}{A_{	ext{initial}}} = \frac{3 	imes 10^{-3}}{3 	imes 10^{-4}} = 10.      $$
The plate separation is unchanged:
$$      \frac{d_{	ext{initial}}}{d_{	ext{new}}} = \frac{3 	imes 10^{-6}}{3 	imes 10^{-6}} = 1.      $$
Overall factor:
$$      10 	imes 1 = 10.      $$
Case E:
New dimensions:
$$l = 5\,	ext{cm} = 0.05\,	ext{m}, \quad b = 2\,	ext{cm} = 0.02\,	ext{m}.$$
New area:
$$      A_{	ext{new}} = 0.05 	imes 0.02 = 1 	imes 10^{-3}\,	ext{m}^2.      $$
Ratio of areas:
$$      \frac{A_{	ext{new}}}{A_{	ext{initial}}} = \frac{1 	imes 10^{-3}}{3 	imes 10^{-4}} \approx 3.33.      $$
New plate separation:
$$d_{	ext{new}} = 1\,\mu	ext{m} = 1 	imes 10^{-6}\,	ext{m},$$
so the ratio of separations is
$$      \frac{d_{	ext{initial}}}{d_{	ext{new}}} = \frac{3 	imes 10^{-6}}{1 	imes 10^{-6}} = 3.      $$
Overall factor:
$$      3.33 	imes 3 \approx 10.      $$
Both Case C and Case E yield a capacitance that is 10 times the original value.
Thus, the correct modifications are those in Case C and Case E.

Answer

C only

Answer

A only

Answer

B and D only

Answer

C and E only

Electrostats & Capacitance

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