Question 197: Electrostatics & Capacitance

Question

A parallel plate capacitor of capacitance 1 µF is charged to a potential difference of 20 V. The distance between plates is 1 µm. The energy density between plates of capacitor is :

Accepted Answer

The energy density ($u$) between the plates of a capacitor is given by the formula:
$$ u = \frac{1}{2} \varepsilon_0 E^2 $$
where $\varepsilon_0$ is the permittivity of free space ($8.85 	imes 10^{-12} \, 	ext{F/m}$) and $E$ is the electric field between the plates. The electric field $E$ is related to the potential difference $V$ and the separation $d$ between the plates by:
$$ E = \frac{V}{d} $$
Given:
Capacitance ($C$) = 1 ÃÂµF = $1 	imes 10^{-6} \, 	ext{F}$
Potential Difference ($V$) = 20 V
Distance ($d$) = 1 ÃÂµm = $1 	imes 10^{-6} \, 	ext{m}$
First, calculate the electric field $E$:
$$ E = \frac{V}{d} = \frac{20 \, 	ext{V}}{1 	imes 10^{-6} \, 	ext{m}} = 2 	imes 10^7 \, 	ext{V/m} $$
Now plug this into the formula for energy density:
$$ u = \frac{1}{2} 	imes 8.85 	imes 10^{-12} \, 	ext{F/m} 	imes (2 	imes 10^7 \, 	ext{V/m})^2 $$
Calculate $u$:
$$ u = \frac{1}{2} 	imes 8.85 	imes 10^{-12} 	imes 4 	imes 10^{14} $$
$$ u = 2 	imes 8.85 	imes 10^2 $$
$$ u = 1770 \, 	ext{J/m}^3 $$
This value is approximately $1.8 	imes 10^3 \, 	ext{J/m}^3$. Therefore, the correct option is:
Option A
: $1.8 	imes 10^3 \, 	ext{J/m}^3$

Answer

$1.8 	imes 10^3$ J/m
3

Answer

$2 	imes 10^2$ J/m
3

Answer

$2 	imes 10^{-4}$ J/m
3

Answer

$1.8 	imes 10^5$ J/m
3

Electrostatics & Capacitance

A parallel plate capacitor of capacitance 1 µF is charged to a potential difference of 20 V. The distance between plates is 1 µm. The energy density between plates of capacitor is :

Choose the correct answer: