Question

A capacitor of capacitance 20 uF is charged to a potential of 500 V.
Calculate the charge and energy stored in a capacitor.
for​

Answer 1

$\huge\underline\red{Given:}$

A capacitor of capacitance, C= 20uF

and it's charged to a potential, V = 500 V

To find:-

Charge

Energy stored in a capacitor

Theory :

A capacitor is a device that stores electrical energy.

• Capacitance :

It is a measure of ability of the capacitor to store charge on it .

• Charge given to the conductor

$\bf\:Q=CVQ=CV$

• Energy Stored in a charged capacitor

$\rm\:U=\dfrac{Q^2}{2C}=\dfrac{1}{2}CV^2=\dfrac{1}{2}$

Solution:

We have :

$\sf\:Capacitance,C=20\mu\:F=20\times10^{-6}$

$\sf\:Potential,V=500VPotential,$

$\sf\:Charge,Q=?$

We know that;

$\rm\red{Charge=Capacitance\times\:Potential}$

$\sf\implies\:Q=CV⟹Q=CV$

$\sf\implies\:Q=20\times10^{-6}$

$\sf\implies\:Q=10\times10^{-3}$

$\sf\implies\:Q=10mC⟹Q=10mC$

Now ,

Energy stored in a capacitor

$\rm\:U=\dfrac{1}{2}$

Put given values then ,

$\sf\implies\:U=\dfrac{1}{2}\times10\times500\times10^{-3}$

$\sf\implies\:U=\dfrac{1}{2}\times10^{3}\times5\times10^{-3}$

$\sf\implies\:U=\dfrac{1}{2}\times10^{(3-3)}$

$\sf\implies\:U=\dfrac{5}{2}$

$\sf\implies\:U=2.5J⟹U=2.5J$

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More About the topic :

• Capacitance is a scalar quantity

SI unit of capacitance is Farad (F)

Dimensions $\sf\:[M^{-1}L^{-2}T^{4}A^{2}]$