 # A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius PEEWSRIGWETRYqx 2021-12-16 Answered

A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge destiny of $+6.37×{10}^{-6}\frac{C}{{m}^{2}}$. A charge of $-0.500\mu C$ is now introduced at the center of the cavity inside the sphere.
(a) What is the new carge density on the outside of the sphere?
(b) Calculate the strength of the electric field just outside the sphere.
(c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?

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(a) Calculation:
Solve for the new charge density on the outside of the sphere:
As equation (22.10) mentions, the electric field at the surface of a conductor is given by
$E=\frac{\sigma }{{ϵ}_{0}}$
We use the same technique as an example (22.5), the electric field at the surface of a charged conducting sphere is given by
$E=\frac{1}{4\pi {ϵ}_{0}}\frac{Q}{{r}^{2}}$
Substituting from the previous calculation and solve for the charge density on the outside of the sphere, then we get
${\sigma }_{out}=E{ϵ}_{0}$
$=\frac{1}{4\pi {ϵ}_{0}}\frac{Q}{{r}^{2}}{ϵ}_{0}$
$=\frac{Q}{4\pi {r}^{2}}$
$=\frac{-0.5×{10}^{-6}C}{4\pi {\left(0.25m\right)}^{2}}$
$=-6.366×{10}^{-7}\frac{C}{{m}^{2}}$ Calculation:
In order to evaluate the new charge density on the outside of the sphere, we use the following relation:
${\sigma }_{\ne w}={\sigma }_{\in }-{\sigma }_{out}$
$=-6.37×{10}^{-6}\frac{C}{{m}^{2}}-6.366×{10}^{-7}\frac{C}{{m}^{2}}$
$=5.733×{10}^{-6}\frac{C}{{m}^{2}}$
So, the new charge density on the outside of the sphere is $5.733×{10}^{-6}\frac{C}{{m}^{2}}$

We have step-by-step solutions for your answer! Becky Harrison
(b) Calculation:
Solve for the strength of the electric field just outside of the sphere:
As equation (22.10) mentions, the electric field at the surface of a conductor is given by
$E=\frac{\sigma }{{ϵ}_{0}}$
$=\frac{5.733×{10}^{-6}\frac{C}{{m}^{2}}}{8.854×{10}^{-12}\frac{{C}^{2}}{N}\cdot {m}^{2}}$
$=6.475×{10}^{5}\frac{N}{C}$
So, the strength of the electric field just outside of the sphere is $6.475×{10}^{5}\frac{N}{C}$

We have step-by-step solutions for your answer! Don Sumner

(c) Calculation:
As equation (22.8) mentions, the electric flux through a spherical surface inside the inner surface of the sphere is given by
${\varphi }_{E}=\frac{{Q}_{enclosed}}{{ϵ}_{0}}$
$=\frac{-0.5×{10}^{-6}C}{8.854×{10}^{-12}{C}^{2}/N\cdot {m}^{2}}$
$=-5.647×{10}^{4}N\cdot {m}^{2}/C$
So, the electric flux through a spherical surface inside the inner surface of the sphere is $-5.647×{10}^{4}N\cdot {m}^{2}/C$

We have step-by-step solutions for your answer!