Question

Question: A thin hollow spherical glass shell of radius carries a uniformly distributed positive charge $+\mathbf{6}\times {\mathbf{10}}^{-\mathbf{9}}\mathbf{C}$, as shown in Figure 15.65. To the right of it is a horizontal permanent dipole with charges $\mathbf{+}\mathbf{3}\times {\mathbf{10}}^{-\mathbf{11}}$and $-\mathbf{3}\times {\mathbf{10}}^{-\mathbf{11}}$separated by a distance (the dipole is shown greatly enlarged for clarity). The dipole is fixed in position and is not free to rotate. The distance from the center of the glass shell to the center of the dipole is 0.6 m.

(a) Calculate the net electric field at the center of the glass shell. (b) If the sphere were a solid metal ball with a charge , what would be the net electric field at its center? (c) Draw the approximate charge distribution in and/or on the metal sphere.

Short answer

Answer

(a) The net electric field at the center of the glass shell is, .

(b) The net electric field at its center is, .

(c) The charge distribution on the sphere has been provided.

Step-by-step solution

Identification of the given data

The given data is listed below as:

• The radius of the thin hollow spherical shell of glass is,$R=0.17\text{m}$
• The charge carried by the thin hollow spherical shell of glass is,$Q=+6\times {10}^{-9}\text{C}$
• The charges of the dipoles are,$q=\pm 3\times {10}^{-11}\text{C}$
• The distance of separation of the dipole, $s=2\times {10}^{-5}\text{m}$
• • The distance between the dipole’s center and the glass shell’s center is, r = 0.6 m

Significance of the electric field 

The electric field of a dipole is directly proportional to the charge and the separation distance of the dipole and inversely proportional to the distance from the center of the dipole to the center of the sphere. Moreover, the electric field of a point charge is directly proportional to the charge of that object and inversely proportional to the distance of the charge from the center of the electric field.

(a) Determination of the net electric field at the center of the glass shell

The equation of the net electric field due to a dipole is expressed as:

$E_1=k\frac{qs}{r^3}$

Here, is the electric field constant, is the charge of the dipole, is the distance of separation of the dipoles and is the distance between the dipole’s center and the glass shell’s center.

Substitute the values in the above equation.

$$\begin{gathered} E_1=\left(9\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(3\times {10}^{-11}\text{C}\right)\left(2\times {10}^{-5}\text{m}\right)}{{\left(0.6\text{m}\right)}^3} \\ =\left(9\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(6\times {10}^{-16}\text{C}·\text{m}\right)}{\left(0.216{\text{m}}^3\right)} \\ =\left(9\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.7\times {10}^{-15}{\text{C/m}}^2\right) \\ =2.5\times {10}^{-5}\text{N/C} \end{gathered}$$

Thus, the net electric field at the center of the glass shell is, $2.5\times {10}^{-5}\text{N/C}$.

(b) Determination of the net electric field at its center

The equation of the net electric field due to the sphere is expressed as:

$E_2=k\frac{Q}{R^2}$

Here, is the electric field constant, is the charge of the sphere and is the radius of the sphere.

Substitute the values in the above equation.

$$\begin{gathered} E_2=\left(9\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(6\times {10}^{-9}\text{C}\right)}{{\left(0.17\text{m}\right)}^2} \\ =\left(9\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(6\times {10}^{-9}\text{C}\right)}{\left(0.0289{\text{m}}^2\right)} \\ =\left(9\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.076\times {10}^{-7}{\text{C/m}}^2\right) \\ =1868.51\text{N/C} \end{gathered}$$

Thus, the net electric field at its center is,1868.51N/C .

(c) Drawing the approximate charge distribution 

The approximate charge distribution has been drawn below:

Here, in this above figure, the charge distribution is at the surface of the sphere. The reason the positive charge is at the surface is that as the positive charge of the dipole is directed at the sphere having positive charges, then due to the repulsion, the positive charge will be outside of the sphere.

Thus, the charge distribution on the sphere has been provided.