Two parallel plate capacitors x and y have the same area of plates and same separation between them. X has air between the plates while y contains a dielectric medium of r = 4 (i) calculate capacitance of each capacitor if equivalent capacitance of the combination is 4 f. (ii) calculate the potential difference between the plates of x and y. (iii) estimate the ratio of electrostatic energy stored in x and y.
Answers
We know that capacitance for a parallel plate capacitor
C= ε₀KA/d
For air, K=1, so capacitance for x, C(x)=ε₀ A/d
Capacitance for y, C(y)=ε₀4K/d= 4C(x)
We know, 1/C= 1/C(x) + 1/C(y)
A/Q, 1/4= 1/C(x) + 1/C(y)
1/4= 1/C(x) + 1/ 4C(x)
1/4= 1/C(x) {1+ 1/4}
= 1/C(x) {5/4}
1/C(x)= 1/5
C(x)= 5 f
So, 1/C(y)= (1/4)- (1/5)
= 1/20
C(y)= 20 f
(ii) We know, Q= CV, where V is the potential difference of the plane.
For, x,
Potential difference V(x) = Q/C(x)
V(x) = Q/5f
For y
Potential difference V(y) = Q/Cy
V(y) = Q/20f
(iii) Now, energy stored in capacitor E= Q²/2C
For E(x)= Q²/2C(x)= Q²/10
For, E(y)= Q²/2C(y) = Q²/40
Ratio of energy, E(x)/E(y)= {(Q²/10)/(Q²/40)}
= 4/1
Answer:
We know that capacitance for a parallel plate capacitor
C= ε₀KA/d
For air, K=1, so capacitance for x, C(x)=ε₀ A/d
Capacitance for y, C(y)=ε₀4K/d= 4C(x)
We know, 1/C= 1/C(x) + 1/C(y)
A/Q, 1/4= 1/C(x) + 1/C(y)
1/4= 1/C(x) + 1/ 4C(x)
1/4= 1/C(x) {1+ 1/4}
= 1/C(x) {5/4}
1/C(x)= 1/5
C(x)= 5 f
So, 1/C(y)= (1/4)- (1/5)
= 1/20
C(y)= 20 f
(ii) We know, Q= CV, where V is the potential difference of the plane.
For, x,
Potential difference V(x) = Q/C(x)
V(x) = Q/5f
For y
Potential difference V(y) = Q/Cy
V(y) = Q/20f
(iii) Now, energy stored in capacitor E= Q²/2C
For E(x)= Q²/2C(x)= Q²/10
For, E(y)= Q²/2C(y) = Q²/40
Ratio of energy, E(x)/E(y)= {(Q²/10)/(Q²/40)}
= 4/1
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