PEEWSRIGWETRYqx

2021-12-16

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?

Ronnie Schechter

(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 $5.733×{10}^{-6}\frac{C}{{m}^{2}}$

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 $6.475×{10}^{5}\frac{N}{C}$

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 $-5.647×{10}^{4}N\cdot {m}^{2}/C$

Eliza Beth13

Step 1:
(a) To find the new charge density on the outside of the sphere, we need to consider the effect of the charge introduced inside the cavity. The charge inside the cavity induces an equal and opposite charge on the inner surface of the sphere. Since the sphere is a conductor, this induced charge redistributes on the outer surface. The new charge density on the outside of the sphere is given by the sum of the original charge density and the induced charge density.
Let $\sigma$ be the original charge density and ${\sigma }_{\text{induced}}$ be the induced charge density. Then the new charge density ${\sigma }_{\text{new}}$ is:
${\sigma }_{\text{new}}=\sigma +{\sigma }_{\text{induced}}$
The induced charge density is equal in magnitude to the charge inside the cavity divided by the inner surface area of the sphere. Therefore:
${\sigma }_{\text{induced}}=\frac{{Q}_{\text{inside}}}{{A}_{\text{inner}}}$ where ${Q}_{\text{inside}}$ is the charge inside the cavity and ${A}_{\text{inner}}$ is the inner surface area of the sphere.
Substituting the given values, we have:
${\sigma }_{\text{induced}}=\frac{-0.500\mu C}{4\pi {r}_{\text{inner}}^{2}}$
Now we can substitute the values and calculate ${\sigma }_{\text{new}}$:
${\sigma }_{\text{new}}=6.37×{10}^{-6}\frac{C}{{m}^{2}}+\frac{-0.500\mu C}{4\pi \left(0.200\phantom{\rule{0.167em}{0ex}}m{\right)}^{2}}$
Step 2:
(b) To calculate the strength of the electric field just outside the sphere, we can use Gauss's law. Gauss's law states that the electric field through a closed surface is proportional to the total charge enclosed by that surface.
The electric field just outside the sphere can be found using the formula:
$E=\frac{{\sigma }_{\text{new}}}{{ϵ}_{0}}$ where ${ϵ}_{0}$ is the vacuum permittivity.
Substituting the values, we have:
$E=\frac{{\sigma }_{\text{new}}}{{ϵ}_{0}}$
Step 3:
(c) The electric flux through a spherical surface just inside the inner surface of the sphere can be calculated using the formula:
$\Phi =\frac{{Q}_{\text{enclosed}}}{{ϵ}_{0}}$ where ${Q}_{\text{enclosed}}$ is the charge enclosed by the surface. In this case, the charge enclosed is the charge inside the cavity.
Substituting the given values, we have:
$\Phi =\frac{{Q}_{\text{inside}}}{{ϵ}_{0}}$

Outer radius of the conducting sphere, ${R}_{\text{outer}}=0.250$ m
Inner radius of the conducting sphere, ${R}_{\text{inner}}=0.200$ m
Surface charge density on the outer surface, $\sigma =+6.37×{10}^{-6}\frac{C}{{m}^{2}}$
Charge introduced inside the cavity, $q=-0.500\mu C=-0.500×{10}^{-6}C$
(a) To find the new charge density on the outside of the sphere, we need to consider the charges on the inner and outer surfaces separately.
The charge on the inner surface of the sphere is equal in magnitude but opposite in sign to the charge introduced inside the cavity. Therefore, the charge on the inner surface, ${Q}_{\text{inner}}=-q=-\left(-0.500×{10}^{-6}C\right)=0.500×{10}^{-6}C$.
The total charge on the outer surface of the sphere, ${Q}_{\text{outer}}$, is the sum of the charge on the outer surface due to surface charge density and the charge on the inner surface due to the charge introduced inside the cavity. Since the inner and outer surfaces are concentric, the total charge on the outer surface is equal to the charge on the inner surface.
Hence, ${Q}_{\text{outer}}={Q}_{\text{inner}}=0.500×{10}^{-6}C$.
The new charge density on the outside of the sphere, ${\sigma }_{\text{new}}$, is given by the ratio of the total charge on the outer surface to the area of the outer surface.
${\sigma }_{\text{new}}=\frac{{Q}_{\text{outer}}}{4\pi {R}_{\text{outer}}^{2}}$
Substituting the given values:
${\sigma }_{\text{new}}=\frac{0.500×{10}^{-6}C}{4\pi \left(0.250\phantom{\rule{0.167em}{0ex}}\text{m}{\right)}^{2}}$
Calculating the value:
${\sigma }_{\text{new}}\approx 1.273×{10}^{-5}\phantom{\rule{0.167em}{0ex}}\frac{C}{{m}^{2}}$
Therefore, the new charge density on the outside of the sphere is $1.273×{10}^{-5}\phantom{\rule{0.167em}{0ex}}\frac{C}{{m}^{2}}$.
(b) To calculate the strength of the electric field just outside the sphere, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space (${\epsilon }_{0}$).
The electric field just outside the sphere is radially outward and has the same magnitude at every point on the surface due to its symmetry. Therefore, we can use a Gaussian surface in the form of a sphere with a radius slightly greater than the outer radius of the conducting sphere.
The electric flux through a closed surface is given by:
${\Phi }_{E}=E·A$ where $E$ is the electric field and $A$ is the area of the closed surface.
The total charge enclosed by the Gaussian surface is ${Q}_{\text{total}}={Q}_{\text{outer}}=0.500×{10}^{-6}C$.
The electric flux through the Gaussian surface is also equal to $\frac{{Q}_{\text{total}}}{{\epsilon }_{0}}$.
Therefore, we have:
$\frac{{Q}_{\text{total}}}{{\epsilon }_{0}}=E·A$
$E=\frac{{Q}_{\text{total}}}{{\epsilon }_{0}A}$
$E=\frac{{Q}_{\text{outer}}}{{\epsilon }_{0}A}$
Substituting the given values:
$E=\frac{0.500×{10}^{-6}C}{{\epsilon }_{0}·4\pi {R}_{\text{outer}}^{2}}$
The area of the closed surface is $A=4\pi {R}_{\text{outer}}^{2}$.
Simplifying further:
$E=\frac{0.500×{10}^{-6}C}{{\epsilon }_{0}·4\pi \left(0.250\phantom{\rule{0.167em}{0ex}}\text{m}{\right)}^{2}}$
Calculating the value:
$E\approx 2.268×{10}^{5}\phantom{\rule{0.167em}{0ex}}\text{N/C}$
Therefore, the strength of the electric field just outside the sphere is $2.268×{10}^{5}\phantom{\rule{0.167em}{0ex}}\text{N/C}$.
(c) To find the electric flux through a spherical surface just inside the inner surface of the conducting sphere, we can use Gauss's law again.
Consider a Gaussian surface in the form of a sphere with a radius slightly less than the inner radius of the conducting sphere.
The electric field inside the conducting sphere is zero since the electric charges redistribute themselves on the inner surface to cancel out the electric field inside.
Therefore, the electric flux through the Gaussian surface is zero, and the electric flux through a spherical surface just inside the inner surface of the sphere is also zero.
In mathematical notation, ${\Phi }_{E}=0$.
Hence, the electric flux through a spherical surface just inside the inner surface of the sphere is $\overline{)0}$.

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