Homework: Electric Potentia Potential of Concentric Spherical Insulator and Cond
ID: 1885004 • Letter: H
Question
Homework: Electric Potentia Potential of Concentric Spherical Insulator and Conductor A solid insulating sphere of radius a = 5.3 cm is fixed at the origin of a co ordinate system as shown. The sphere is uniformly charged with a charge density =-153 C/m3 Concentric with the sphere is an uncharged spherical conducting shell of inner radius b 14.6 cm, and outer radius c 16.6 cm. Pan What is Ex(P), the x-component of the electric field at point P located a distance d-26 cm from the origin along the x-axis as shown? What is V(b), the electric potential at the inner surface of the conducting shell? Define the potential to be zero at infinity. What is V(a), the electric potential at the outer surface of the insulating sphere? Define the potential to be zero at infinity. "What is V(c) - V(a), the potentital differnece between the outer surface of the conductor and the outer surface of the insulator? 5) A charge Q . 0.0205C is now added to the conducting shell. What is V(a), the electric potential at the outer surface of the insulating sphere, now? Define the potential to be zero at infinity. "Below is some space to write notes on this problem.Explanation / Answer
given
solid insulating sphere
radius a = 5.3 cm
rho = -153 uC/m^3
b = 14.6 cm
c = 16.6 cm
a. at d = 25 cm
E(x) = k(4*pi*rho*a^3/3)/d^2
E(x) = -0.153*8.98*4*pi*0.053^3/3*0.25^2
E(x) = 0.013708942922 N/C
b. V(b) = k(4*pi*rho*a^3/3)/b
V(b) = -8.98*0.153*4*p*0.053^3/3*0.146 = -0.0052244447114 V
c. V(a) = k(4*pi*rho*a^3/3)/a
hence
V(a) = k*4*pi*rho*a^2/3
V(a) = -0.0161662062 V
d. V(c) - V(a) = ?
V(c) = V(b) ( because potential change inside the conductor fropm b to c will be 0)
V(c) - V(a) = -0.0052244447114 + 0.0161662062 = 0.010941761488 V
e. Q = 0.0205 uC
V(a) = -0.0161662062 + kQ/c = -0.01505723029638 V
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