Chapter 16, Problem 67P

(a)

To determine

The magnitude of the electric field at a point $1.75\text{cm}$ from the center.

Answer to Problem 67P

The magnitude of the electric field at a point $1.75\text{cm}$ from the center is $\underset{\_}{6.8\times {10}^6\text{N/C}}$.

Explanation of Solution

Given that the charge of the inner sphere is $230\text{nC}$, the distance to the point under consideration is $1.75\text{cm}$, The inner radius of the shell is $2.25\text{cm}$, the outer radius of the shell is $2.75\text{cm}$, and the radius of the inner sphere is $1.50\text{cm}$.

The point under consideration is in between the inner radius of the shell and the sphere. The electric field between the charged sphere and the spherical shell is the same as that due to a point charge at the center of the sphere with same charge as the sphere.

Write the expression for the magnitude of electric field at an outside point of a charged sphere.

$E=\frac{kq}{r^2}$ (I)

Here, $E$ is the magnitude of electric field, $k$ is a constant ($k=8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$), $q$ is the charge of the sphere, and $r$ is the distance to the point from the center of the sphere.

Conclusion:

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $230\text{nC}$ for $q$, $1.75\text{cm}$ for $r$ in equation (I) to find $E$.

$\begin{array}{c}E=\frac{\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(230\text{nC}\right)}{{\left(1.75\text{cm}\right)}^2}\\ =\frac{\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(230\text{nC}\times \frac{1\text{C}}{1\times {10}^9\text{nC}}\right)}{{\left(1.75\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =6.8\times {10}^6\text{N/C}\end{array}$

Therefore, the magnitude of the electric field at a point $1.75\text{cm}$ from the center is $\underset{\_}{6.8\times {10}^6\text{N/C}}$.

(b)

To determine

The magnitude of the electric field at a point $2.50\text{cm}$ from the center.

Answer to Problem 67P

The magnitude of the electric field at a point $2.50\text{cm}$ from the center is $\underset{\_}{\text{zero}}$.

Explanation of Solution

Given that the charge of the inner sphere is $230\text{nC}$, the distance to the point under consideration is $2.50\text{cm}$, The inner radius of the shell is $2.25\text{cm}$, the outer radius of the shell is $2.75\text{cm}$, and the radius of the inner sphere is $1.50\text{cm}$.

The electric field inside any conducting material in electrostatic equilibrium is zero. The point under consideration in inside the conducting spherical shell. This indicates that the electric field at any point inside the conducting spherical shell is zero.

Conclusion:

Therefore, the magnitude of the electric field at a point $2.50\text{cm}$ from the center is $\underset{\_}{\text{zero}}$.

(c)

To determine

The electric field at a point $3.00\text{cm}$ from the center.

Answer to Problem 67P

The electric field at a point $3.00\text{cm}$ from the center is $\underset{\_}{2.3\times {10}^6\text{N/C, directed away from the center}}$.

Explanation of Solution

Given that the charge of the inner sphere is $230\text{nC}$, the distance to the point under consideration is $3.00\text{cm}$, The inner radius of the shell is $2.25\text{cm}$, the outer radius of the shell is $2.75\text{cm}$, and the radius of the inner sphere is $1.50\text{cm}$.

The point under consideration outside the sphere as well as shell. The electric field outside of the spherical shell enclosing the charged sphere is the same as that due to a point charge at the center of the sphere with same charge as the sphere.

Equation (I) gives the expression for the magnitude of electric field at an outside point of a charged sphere.

$E=\frac{kq}{r^2}$

Conclusion:

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $230\text{nC}$ for $q$, $3.00\text{cm}$ for $r$ in equation (I) to find $E$.

$\begin{array}{c}E=\frac{\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(230\text{nC}\right)}{{\left(3.00\text{cm}\right)}^2}\\ =\frac{\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(230\text{nC}\times \frac{1\text{C}}{1\times {10}^9\text{nC}}\right)}{{\left(3.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =2.3\times {10}^6\text{N/C}\end{array}$

Electric field from a positive point charge is always diverges away from it. For a sphere of positive charge, the electric field will be radially outward in direction. Hence, here, the electric field is directed away from the center.

Therefore, the electric field at a point $3.00\text{cm}$ from the center is $\underset{\_}{2.3\times {10}^6\text{N/C, directed away from the center}}$.