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Question:
A spherical capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 10.0 cm and the outer sphere has radius 14.5 cm. A potential difference of 110 V is applied to the capacitor.
(a) What is the energy density at r = 10.1 cm, just outside the inner sphere? in J/m3
(b) What is the energy density at r = 14.4 cm, just inside the outer sphere? in J/m3
Answer:
U = 0.0218 j/m³
U = 0.00527 j/m³
Given Information:
Radius inner sphere = ri = 10 cm = 0.10 m
Radius outer sphere = ro = 14.5 cm = 0.145 m
Potential difference = V = 110 V
E = 8.85x10⁻¹² C²/N.m²
Required Information:
Energy density U = at r = 10.1 cm = 0.101 m
Energy density U = at r = 14.4 cm = 0.144 m
Step-by-step explanation:
(a) What is the energy density at r = 10.1 cm, just outside the inner sphere?
The capacitance can be found by
Capacitance = 4πEo(ri*ro/ri+ro)
Capacitance = 4π8.85x10⁻¹²(0.10*0.145/0.10+0.145)
Capacitance = 6.581x10⁻¹² F
The charge on the capacitor can found using
Charge = Q = CV
Q = 6.581x10⁻¹²*110 = 7.964x10⁻⁸
The Electiric field is given by
E = 1/4πEo(Q/r²)
where r = 0.101 m
E = 1/4π8.85x10⁻¹²(7.964x10⁻⁸/(0.101²)
E = 70199.7 V/m
Energy density is given by
U = 1/2EoE²
U = 1/2*8.85x10⁻¹²(70199.7)²
U = 0.0218 j/m³
(b) What is the energy density at r = 14.4 cm, just inside the outer sphere?
The Electiric field is given by
E = 1/4πEo(Q/r²)
where r = 0.144 m
E = 1/4π8.85x10⁻¹²(7.964x10⁻⁸/(0.144²)
E = 34534.5 V/m
Energy density is given by
U = 1/2EoE²
U = 1/2*8.85x10⁻¹²(34534.5)²
U = 0.00527 j/m³