Question

A certain parallel-plate capacitor is filled with a dielectric for which κ = 5.0. The area of each plate is 0.027 m2, and the plates are separated by 2.0 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 200 kN/C. What is the maximum energy that can be stored in the capacitor?

Answer 1

Energy will be
`4.776* 10^(-5)J`

Step-by-step explanation:

We have given that dielectric constant k = 5

Area of the plate
`A=0.027m^2`

Distance between the plates
`d=2mm=2* 10^(-3)m`

Electric field E = 200 kN/C

We know that capacitance is given by
`C=(K\epsilon _0A)/(d)=(5* 8.85* 10^(-12)* 0.027)/(2* 10^(-3))=0.597* 10^(-9)F`

We know that electric field is given by
`E=(V)/(d)`

So
`V=Ed=200* 10^3* 2* 10^(-3)=400volt`

So energy will be
`E=(1)/(2)CV^2=(1)/(2)* 0.597* 10^(-9)* 400^2=4.776* 10^(-5)J`