1) For a parallel plate capacitor with Area A and plate separation d a. Write th
ID: 1847388 • Letter: 1
Question
1) For a parallel plate capacitor with Area A and plate separation d
a. Write the total Energy stored only in terms of the E-field and again only in terms of the total charge (and dimensions); the dielectric is just air i.e. e=eo
b. Describe and explain what happens to the stored Energy when you insert a dielectric medium with dielectric constant e=ereo. The capacitor had been initially charged with charge Q and is not connected to an external circuit. Also describe how the field and charge changes ( or doesn
Explanation / Answer
E = electric field = Q/(A*e )
V = potentail difference between plate = E*d
C = capacitance = Q/V = Q/( E*d )
Energy = C*V^2/2 = [Q/(Ed) ] (Ed)^2 /2 = (Q*E*d)/2
b) when dielectric is inserted , E' = E/er
V' = E' *d = E*d/er = Ed/er = V/er
Q' = Q/er = charge on capaciotr will decrsased , by a factor of er
Then C' = Q/V' = er *Q/( E*d ) = er* C
Energy = C' * V' ^2 /2
= (er C )* ( V/er ) ^2 /2
= er ( C V^2 ) /[ 2*er^2 ]
= CV^2 / (2 *er )
= (Q*E*d)/(2er)
Hence stored energy is dereased by a factor of er
when capcitor is charged to Q , then charge on it remains same after insertion of dielectric ,
E = electric field = Q/ ( A*e*er )
so electric field will decreased by a factor of er
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