Question

The electric field at point Pjust outside the outer surface of a hollow spherical conductor of inner radius 10 cmand outer radius 20 cmhas magnitude 450 N/ Cand is directed outward. When a particle of unknown charge Qis introduced into the center of the sphere, the electric field at Pis still directed outward but is now 180 N/C.

(a) What was the net charge enclosed by the outer surface before Qwas introduced?

(b) What is charge Q?

After Qis introduced, what is the charge on the

(c) inner and

(d) outer surface of the conductor?

Short answer

a) Net charge enclosed by the outer surface before Q is introduced is$2.0\times {10}^{-9}C$ .

b) The value of charge Q is $-1.2\times {10}^{-9}C$.

c) Charge on the inner surface of the conductor is $-1.2\times {10}^{-9}C$.

d) Charge on the outer surface of the conductor is$+0.80\times {10}^{-9}C$ .

Step-by-step solution

Listing the given quantities

• A hollow sphere has an inner radius of r = 10 cm and an outer radius of R=20 cm.
• Electric field, ${\mathrm{E}}_{\mathrm{initial}}=450\mathrm{N}/\mathrm{C}$(when the charge is absent).
• Electric field, ${\mathrm{E}}_{\mathrm{final}}=180\mathrm{N}/\mathrm{C}$(When the charge is present inside).

Understanding the concept of charge density and electric field

The initial field (evaluated “just outside the outer surface,” which means it is evaluated at=${\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{m}$, the outer radius of the conductor) is related to the charge q on the hollow conductor by .

${\mathbf{E}}_{\mathbf{initial}}\mathbf{=}\mathbf{q}\mathbf{/}\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}{\mathbf{R}}_{\mathbf{2}}^{\mathbf{2}}$

After the point charge Q is placed at the geometric center of the hollow conductor, the final field at that point is a combination of the initial and that due to Q.

(a) Calculation of the net charge enclosed by the outer surface

The electric field due to a charge on the spherical shell outside the shell is given by the formula,

$E\frac{q}{{\epsilon }_{0}4{\mathrm{\pi r}}^2}$

Here, E is the electric field, is the enclosed charge, and r is the radius of the shell $E_{initial}$.

Now, let’s assume that the radius of the shell is R and the initial electric field on the shell is . Therefore, we can write,

$$\begin{gathered} E_{initial}=\frac{q}{{\epsilon }_{0}4{\mathrm{\pi R}}^2} \\ q=E_{initial}{\epsilon }_{0}4{\mathrm{\pi R}}^2 \end{gathered}$$

Now, substitute the given values in the above equation to find the charge.

$$\begin{gathered} \mathrm{q}=\frac{\left(450\mathrm{N}/\mathrm{C}\right){\left(0.02\mathrm{m}\right)}^2}{9.0\times {10}^9\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}} \\ =2.0\times {10}^{-9}\mathrm{C} \end{gathered}$$

Net charge enclosed by the outer surface before Q is introduced is .

(b) Calculation of the value of charge Q

After the charge Q is introduced, the electric field is changed. The value of the newly introduced charge can be found using the difference in the values of initial and final electric fields.

$$\begin{gathered} Q=4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2.\left({\mathrm{E}}_{\mathrm{final}}-{\mathrm{E}}_{\mathrm{initial}}\right) \\ =\frac{{\left(0.02\mathrm{m}\right)}^2\left(180\mathrm{N}/\mathrm{C}-450\mathrm{N}/\mathrm{C}\right)}{9.0\times {10}^9\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}} \\ =-1.2\times {10}^{-9}\mathrm{C} \end{gathered}$$

The value of charge Q is .

(c) Calculation of the charge on the inner surface of the conductor

In order to cancel the field (due to Q) within the conducting material, there must be an amount of charge equal to –Q distributed uniformly on the inner surface (radius ). Thus, the answer is $1.2\times {10}^{-9}C$.

Charge on the inner surface of the conductor is $1.2\times {10}^{-9}C$.

(d) Calculation of the charge on the outer surface of the conductor

Since the total excess charge on the conductor is q and is located on the surfaces, then the outer surface charge must equal the total minus the inner surface charge. Thus, the answer is:

$$\begin{gathered} =2.0\times {10}^{-9}\mathrm{C}-1.2\times {10}^{-9}\mathrm{C} \\ =+0.80\times {10}^{-9}\mathrm{C} \end{gathered}$$

Thus, the charge on the outer surface of the conductor is $=+0.80\times {10}^{-9}\mathrm{C}$.