Question

The two plates of a parallel plate capacitor of same lengths have unequal areas as $100{\text{cm}}^2$ and $120{\text{cm}}^2$ respectively. If the separation between the plates is $10\text{mm}$ throughout, then its capacitance will be

• A. $17.70\text{pF}$
• B. $8.85\text{pF}$
• C. $4.425\text{pF}$
• D. $2.25\text{pF}$

Answer 1

The correct option is B $8.85\text{pF}$

We know that, capacitance of a parallel plate capacitor is $$C=\frac{{\epsilon }_{0}A}{d}$$ Where, $A$ is the effective area and $d$ is the separation between the plates.

When, the plates have an unequal area, then the individual area does not matter.

Then the effective area of overlap is considered to calculate the capacitance, which is decided by the plate with the smaller area.

In this case, effective area, $A=100{\text{cm}}^2$ and $d=10\text{mm}$

Thus,

$C=\frac{8.85\times {10}^{-12}\times (100\times {10}^{-4})}{10\times {10}^{-3}}$

$\Rightarrow C=8.85\times {10}^{-12}\text{F}$

$\Rightarrow C=8.85\text{pF}$

Hence, option $\left(\text{B}\right)$ is the correct answer.