Question

There is a parallel plate capacitor with capacitance C=1.4X10^-12 F . If a dielectric material with dielectric constant 2.8 is placed between plates, what is energy stored in it under a potential difference 1500 V?

Answer 1

Given about parallel plate capacitor :-

Capacitance (C)= 1.4×10^-12 F

Dielectric constant (K) = 2.8

Potential difference (V) = 1500 V

Note :- when dielectric material with dielectric constant k is placed between parallel plates then capacitance increases by K factor and energy stored in capacitor is also increases by K factor .

​​​​​​

​​​​New Capacitance (C') = K*C

C' = 2.8*(1.4×10^-12 F )

C' = 3.92×10^-12 F

Formulae:-

Energy stored in new capacitor E' = 0.5*(C')*(V²)

E' = 0.5*(3.92×10^-12)*(1500)²

E'= 0.5*(3.92×10^-12)*(2.25×10⁶)

E' = 0.5*8.82×10^-6

E' = 4.41×10^-6

E' = 4.41 micro joules (or) J

Hence energy stored in new capacitor after adding dielectric material is E' = 4.41 J