Question

The area of each plate of a parallel plate capacitor is 0.08m2. The distance between the plates of the capacitor is 2×10-3m, and the potential difference between the plates is 150V. Calculate capacitance of the capacitor
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kindly solve this with explain

Answer 1

Given:

In parallel plate capacitor

Area of plates, A= 0.08 m²= 8×10⁻² m²

Distance between plates, d= 2×10⁻³m

Potential difference between plates= 150V

To Find:

Capacitance of capacitor

Solution:

We know that,

• Capacitance C of parallel plate capacitor is given by

$\pink{\boxed{\bf{C=\frac{A\epsilon_{\circ}}{d}}}}$

where,

A is the area of the plates

∈₀ is the permittivity of air

∈₀ = 8.854×10⁻¹² C²/Nm²

d is the distance between plates

$\rule{190}{1}$

Let the capacitance of given parallel plate capacitor be C

So,

$\longrightarrow\rm{C=\dfrac{A\epsilon_{\circ}}{d}}$

$\longrightarrow\rm{C=\dfrac{\cancel{8}\times10^{-2}\times8.854\times10^{-12}}{\cancel{2}\times10^{-3}}}$

$\longrightarrow\rm{C=\dfrac{4\times10^{-2}\times8.854\times10^{-12}}{10^{-3}}}$

$\longrightarrow\rm{C=\dfrac{35.416\times10^{-2}\times10^{-12}}{10^{-3}}}$

$\longrightarrow\rm{C=\dfrac{35.416\times10^{-14}}{10^{-3}}}$

$\longrightarrow\rm\green{C=35.416\times10^{-11}\:F}$

Hence, the capacitance of given capacitor is 35.416×10⁻¹¹F.