A 32 cm -diameter conducting sphere is charged to 530 V relative to V=0 at r=∞? Part A What is the surface charge density σ? Express your answer using two significant figures.Part B At...

A 32 cm -diameter conducting sphere is charged to 530 V relative to V=0 at r=∞?

Part A
What is the surface charge density σ?
Express your answer using two significant figures.


Part B
At what distance from the center of the sphere will the potential due to the sphere be only 21 V ?
Express your answer using two significant figures.

Asked on by yikulah

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electreto05 | College Teacher | (Level 1) Assistant Educator

Posted on

a)

In the outer points to a charged sphere the electric field behaves as if all the load were at the center of the sphere, so that the surface potential can be expressed as the potential created by an electrical charge Q at a distance r, equal to the radius of the sphere. The potential of a particle has the following expression:

V = Q/4πԐ0r        (1)

Where:

Q, is the electric charge.

Ԑ0, is the electric permittivity of vacuum.

r, is the distance

The electric charge Q is equal to the charge density σ, multiplied by the surface of the sphere:

Q = σ*S = σ*(4πr^2)

Substituting into the potential equation, we have:

V = (σ*4πr^2)/4πԐ0r = σ*r/Ԑ0       

σ = V*Ԑ0/r = (530)(8.854*10^-12)/(0.16)

σ = 2.93*10^-8 C/m^2

b)

With the charge density, we can calculate the electric charge:

Q = σ*S = σ(4πr^2) = (2.93*10^-8)(12.56)(0.0256)

Q = 9.42*10^-9 C

Applying the equation (1)

r = Q/4πԐ0V = (9.42*10^-9)/(12.56)(8.854*10^-12)(21)

r = 4.03 m

r = 4.03 m, from the center of the sphere.     

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