In the figure,the charge on the capacitor is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The energy stored in a capacitor is given by the formula:$U = \frac{1}{2} C V^{2} \dots (i)$where $C$ is the capacitance and $V$ is the potential

Vedclass · Accepted Answer

$12 \mu F, 600 \mu J$

Vedclass · Answer

(B) The energy stored in a capacitor is given by the formula:$U = \frac{1}{2} C V^{2} \dots (i)$where $C$ is the capacitance and $V$ is the potential difference.From the given graph,the slope is:$	ext{Slope} = 	an 	heta = \frac{Q}{V} = \frac{120 \mu C}{10 	ext{ V}} = 12 \mu F \dots (ii)$Since the capacitance $C = \frac{Q}{V}$,we have:$C = 12 \mu F = 12 	imes 10^{-6} 	ext{ F}$Now,substituting the values of $C = 12 	imes 10^{-6} 	ext{ F}$ and $V = 10 	ext{ V}$ into Eq. $(i)$:$U = \frac{1}{2} 	imes (12 	imes 10^{-6} 	ext{ F}) 	imes (10 	ext{ V})^{2}$$U = \frac{1}{2} 	imes 12 	imes 10^{-6} 	imes 100$$U = 6 	imes 10^{-4} 	ext{ J} = 600 	imes 10^{-6} 	ext{ J} = 600 \mu J$Thus,the capacitance is $12 \mu F$ and the energy stored is $600 \mu J$.

In the figure,the charge on the capacitor is plotted against the potential difference across the capacitor. The capacitance and energy stored in the capacitor are respectively:

Similar Questions

In steady state,the energy stored in the capacitor as shown in the figure is ..... $nJ$.

If $E$ is the electric field intensity of an electrostatic field,then the electrostatic energy density is proportional to

If potential difference across a capacitor is changed from $15 \ V$ to $30 \ V,$ work done is $W.$ The work done,when potential difference is changed from $30 \ V$ to $60 \ V,$ will be $.......W$

$A$ capacitor of $6 \ \mu F$ is charged from $10 \ V$ to $20 \ V$. The increase in its stored energy is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module