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The plates of a parallel plate capacitor...

The plates of a parallel plate capacitor have an area of ` 100 cm^(2)` each and are separated by 3 mm . The capacitor is charged by connecting it to a 400 V supply .
If a dieletric of dielectric constant 2.5 is introduced between the plates of the capacitor , then find the electrostatic energy stored and also change in the energy stored .

Text Solution

Verified by Experts

Given ` A = 100 cm^(2) = 10^(2) xx 10^(-4) = 10^(-2) m^(2)`
d = 3 mm = ` 3xx 10^(-3) m`,
V = 400 V
` E^(1) = in _(r) E = 2.5 xx 2.61 xx 10^(-6) = 5.903 xx 10^(-6) J`
`Delta E = E^(1) - E = (5.903 - 2.361 ) xx 10^(-6) J `
Difference in the energy stored ` Delta E ` ,
` Delta E = 3.542 xx 10^(-6) J `
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