Question

An ideal parallel plate capacitor with a cross-sectional area of 0.4 cm2 contains a dielectric with a dielectric constant of 4 and a dielectric strength of 2 x 108 V/m. The separation between the plates of the capacitor is 5 mm. What is the maximum electric charge (in nC) that can be stored in the capacitor before dielectric breakdown

Answer 1

The maximum charge is 7.08 x 10^-8 C.

Step-by-step explanation:

Area, A = 0.4 cm^2

K = 4

Electric field, E = 2 x 10^8 V/m

separation, d = 5 mm = 0.005 m

Let the capacitance is C and the charge is q.

`q = CV\\\\q=(\varepsilon o A)/(d)* E d\\\\q = \varepsilon o A E\\\\q = 8.85* 10^(-12)*0.4* 10^(-4)* 2* 10^8\\\\q = 7.08* 10^(-8)C`

Answer 2

Answer:
`283.2* 10^(-9)\ nC`

Step-by-step explanation:

Given

Cross-sectional area
`A=0.4\ cm^2`

Dielectric constant
`k=4`

Dielectric strength
`E=2* 10^8\ V/m`

Distance between capacitors
`d=5\ mm`

Maximum charge that can be stored before dielectric breakdown is given by

`\Rightarrow Q=CV\\\\\Rightarrow Q=(k\epsilon_oA)/(d)\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4* 8.85* 10^(-12)* 0.4* 10^(-4)* 2* 10^8\\\\\Rightarrow Q=28.32* 10^(-8)\\\\\Rightarrow Q=283.2* 10^(-9)\ nC`