91影视

• 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).

The initial field (evaluated 鈥渏ust outside the outer surface,鈥� 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 .

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

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.

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

$E\frac{q}{{蔚}_{0}4{\mathrm{蟺谤}}^2}$

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

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

$$\begin{gathered} E_{initial}=\frac{q}{{蔚}_{0}4{\mathrm{蟺搁}}^2} \\ q=E_{initial}{蔚}_{0}4{\mathrm{蟺搁}}^2 \end{gathered}$$

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

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

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

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.

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

The value of charge Q is .

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

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

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:

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

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