A solid insulating sphere of radius a = 4.7 cm is fixed at the origin of a co-or

Question

A solid insulating sphere of radius a = 4.7 cm is fixed at the origin of a co-ordinate system as shown. The sphere is uniformly charged with a charge density ? = -336 ?C/m3. Concentric with the sphere is an uncharged spherical conducting shell of inner radius b = 11.2 cm, and outer radius c = 13.2 cm.

Potential of Concentric Spherical Insulator and Conductor 1 2 3 4 5 6 A solid insulating sphere of radius a = 4.7 cm is fixed at the origin of a co-ordinate system as shown. The sphere is uniformly charged with a charge density =-336 C/m Concentric with the sphere is an uncharged spherical conducting shell of inner radius b 11.2 cm, and outer radius c = 13.2 cm. P(a,0) 1) What is Ex(P), the x-component of the electric field at point P, located a distance d - 39 cm from the origin along the x-axis as shown? N/C Submit 2) What is V(b), the electric potential at the inner surface of the conducting shell? Define the potential to be zero at infinity. Submit 3) What is V(a), the electric potential at the outer surface of the insulating sphere? Define the potential to be zero at infinity. Submit 4) What is V(c) - V(a), the potentital differnece between the outer surface of the conductor and the outer surface of the insulator? Submit 5) A charge Q = 0.0259pC 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. Submit

Explanation / Answer

Here ,

p = -336 uC/m^3

1) for the charge on the inner sphere ,

Q1 = volume * charge density

Q1 = - 4pi/3 * 0.047^3 * 336 uC

Q1 = -0.146 uC

now , at the electric field at Ex(P)

Ex(P) = -9*10^9 * (0.146 * 10^-6)/.39^2

Ex(P) = -8639 N/C

the x-component of electric field is - 8639 N/C

2)

at the inner surface of conducting shell ,

potential = k * Q1/c

potential = - 9*10^9 * 0.146 *10^-6/132

potential = - 9954 V

the potential at point P is - 9954 V

3)

at the outer surface of insulating surface ,

potential = k * Q1/a

potential = - 9*10^9 * 0.146 *10^-6/0.047

potential = - 27960 V

the potential at outer surface of insulating surface is 27960 V

4)
Potential difference , V(c) - V(a) = -9954 - (-27960 )

V(c) - V(a) = 18003 V

the potential difference V(c) - V(a) is 18003 V

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.