OpenResponse Homework Problem 71 A spherical capacitor is fo

Open-Response Homework Problem 7.1 A spherical capacitor is formed from an inner conducting sphere of radius a = 10cm, a dielectric shell with inner radius b = 15cm and outer radius c = 20cm, and an outer conducting shell with inner radius d = 25cm. The dielectric shell has dielectric constant ? = 3. For the computation of the capacitance, assume an arbitrary charge of Q on the inner conductor and ?Q on the outer conductor. (a)Compute the potential difference across region I, ?VI , in terms of Q. (b)Compute the potential difference across region II, ?VII , in terms of Q. (c)Compute the potential difference across region III, ?VIII , in terms of Q. (d)Compute the capacitance, both symbolically and numerically.

Part 2

A conducting sphere of radius a = 5cm is surrounded by a dielectric shell, with constant dielectric ? = 5, inner radius b = 10cm, and outer radius c = 15cm. The system is shown to the right. (a)Compute the capacitance between the conductor and ground. Report both the symbolic value and a numeric value. Clearly report and label any intermediate quantities used to get your result and carefully show your work. (b)If the conductor has a potential difference of 100V with the ground, what charge is on the conductor? (c)What is the energy of the system if the potential difference between the inner conductor and ground is 100V?

Open-Response Homework Problem 7.1 A spherical ca- pacitor is formed from an inner conducting sphere of radius a - 10cm, a dielectric shell with inner radius b - 15cm and outer radius c - 20cm, and an outer conducting shell with inner radius d- 25cm. The dielectric shell has dielectric con- stant K - 3. For the computation of the capacitance, assume an arbitrary charge of Q on the inner conductor and -Q on Conductor Air Dielectric Air the outer conductor. Conductor (a)Compute the potential difference across region I, (b)Compute the potential difference across region II, (c)Compute the potential difference across region IlI, d)Compute the capacitance, both symbolically and nu- VI, in terms of Q AV VI, in terms of Q AV, in ter VIII, in terms of Q merically

Solution

charge on inner conductor =q

Electric field in the region I E = kq/r²

                                             =9.0e+9 q/r²

potential in the region I is given by

V_I = ò9.0e+9 q/r² with limits r=0.1 to 0.15

=          - 9.0e+9 q/r   , put limits

= 9.0e+9q *(1/0.1 -1/0.15)

= 30.0e+9 q V

Electric field in the region II E = kq/Kr²

                                             =3.0e+9 q/r²

potential in the region II is given by

V_II = ò3.0e+9 q/r² with limits r=0.15 to 0.20

=          - 3.0e+9 q/r   , put limits

= 3.0e+9q *(1/0.15 -1/0.20)

= 5.0e+9 q V

Electric field in the region III E = kq/r²

                                             =9.0e+9 q/r²

potential in the region II is given by

V_III = ò9.0e+9 q/r² with limits r=0.20 to 0.25

=          - 9.0e+9 q/r   , put limits

= 9.0e+9q *(1/0.20 -1/0.25)

= 9.0e+9 q V

Potential difference between the 2 conducting shells

V = 30.0e+9 q – 9.0e+9 q = 21.0e+9q

capacitance C = q/V = 1/21.0e+9

                        = 4.76e-11 F