Chapter 23, Problem 9P

(a) Consider a uniformly charged, thin-walled, right circular cylindrical shell having total charge Q, radius R, and length ℓ. Determine the electric field at a point a distance d from the right side of the cylinder as shown in Figure P23.9. Suggestion: Use the result of Example 23.2 and treat the cylinder as a collection of ring charges. (b) What If? Consider now a solid cylinder with the same dimensions and carrying the same charge, uniformly distributed through its volume. Use the result of Example 23.3 to find the field it creates at the same point.

Figure P23.9

To determine

The electric field at a point a distance $d$ from the right side of the thin walled right circular cylindrical shell.

Answer to Problem 9P

The electric field at a point a distance $d$ from the right side of the thin walled right circular cylindrical shell is $\frac{k_{e}Q\widehat{i}}{h}\left[\frac{1}{{\left(d^2+R^2\right)}^{1/2}}-\frac{1}{{\left({\left(d+h\right)}^2+R^2\right)}^{1/2}}\right]$.

Explanation of Solution

Given info: The total charge is $Q$, the radius of the cylindrical shell is $R$, the height of the cylindrical shell is $h$, the distance of point from the right side of the cylinder is $d$.

Consider one ring with thickness $dx$ and it has charge $\frac{Qdx}{h}$.

Figure (1)

The formula used to find the electric field at the chosen point is,

$dE=\frac{k_{e}x}{{\left(x^2+R^2\right)}^{\frac{3}{2}}}\frac{Qdx}{h}\widehat{i}$

Here

$x$ is the point where electric field is to be find out.

$k_e$ is the electrostatic constant.

Integrate on both side from the limits $d$ to $d+h$

${\displaystyle \int dE={\displaystyle \underset{d}{\overset{d+h}{\int }}\frac{k_{e}Qxdx}{l{\left(x^2+R^2\right)}^{3/2}}\widehat{i}}}$,

$\begin{array}{c}E={\displaystyle \underset{d}{\overset{d+h}{\int }}\frac{k_{e}Qdx}{h{\left(x^2+R^2\right)}^{3/2}}\widehat{i}}\\ =\frac{k_{e}Q}{2h}{\displaystyle \underset{d}{\overset{d+h}{\int }}{\left(x^2+R^2\right)}^{-3/2}2xdx}\widehat{i}\\ =\frac{k_{e}Q}{2h}{\left[\frac{\left(x^2+R^2\right)}{\left(-1/2\right)}\right]}_d^{d+h}\widehat{i}\\ =\frac{k_{e}Q\widehat{i}}{h}\left[\frac{1}{{\left(d^2+R^2\right)}^{1/2}}-\frac{1}{{\left({\left(d+h\right)}^2+R^2\right)}^{1/2}}\right]\end{array}$

Conclusion:

Therefore, the electric field at a point a distance $d$ from the right side of the thin walled right circular cylindrical shell is $\frac{k_{e}Q\widehat{i}}{h}\left[\frac{1}{{\left(d^2+R^2\right)}^{1/2}}-\frac{1}{{\left({\left(d+h\right)}^2+R^2\right)}^{1/2}}\right]$.

To determine

The electric field created at the same point by a solid cylinder with same dimensions and carrying the same charge.

Answer to Problem 9P

The electric field created at the same point by a solid cylinder with same dimensions and carrying the same charge is $\frac{2k_{e}Q\widehat{i}}{R^{2}h}\left[h+{\left(d^2+R^2\right)}^{1/2}-{\left({\left(d+h\right)}^2+R^2\right)}^{1/2}\right]$

Explanation of Solution

Given info: The total charge is $Q$, the radius of the solid cylinder is $R$, the height of the solid cylinder is $h$, the distance of point from the right side of the cylinder is $d$.

The charge per unit area is,

$\sigma =\frac{Qdx}{\pi R^{2}h}$

Here,

$\sigma$ is the surface charge density.

The field produced by the one disk is,

$dE=\frac{2\pi k_{e}Qdx}{\pi R^{2}h}\left(1-\frac{x}{{\left(x^2+R^2\right)}^{1/2}}\right)\widehat{i}$

Here,

$x$ is the point where electric field is to be find out.

$k_e$ is the electrostatic constant

Integrate on the both side to get the total electric field

$\begin{array}{c}E={\displaystyle {\int }_d^{d+h}\frac{2\pi k_{e}Qdx}{\pi R^{2}h}\left(1-\frac{x}{{\left(x^2+R^2\right)}^{1/2}}\right)\widehat{i}}\\ =\frac{2k_{e}Q}{R^{2}h}\left[{\displaystyle {\int }_d^{d+h}dx-\frac{1}{2}d+{\displaystyle {\int }_d^{d+h}{\left(x^2+R^2\right)}^{-1/2}2xdx}}\right]\widehat{i}\\ =\frac{2k_{e}Q}{R^{2}h}\left[{\displaystyle {\int }_d^{d+h}dx-\frac{1}{2}{\displaystyle {\int }_d^{d+h}{\left(x^2+R^2\right)}^{-1/2}2xdx}}\right]\widehat{i}\\ =\frac{2k_{e}Q}{R^{2}h}\left[{\left(x\right)}_d^{d+h}-\frac{1}{2}{\left(\frac{{\left(x^2+R^2\right)}^{1/2}}{1/2}\right)}_d^{d+h}\right]\end{array}$

Further solve the above expression.

$\begin{array}{c}E=\frac{2k_{e}Q}{R^{2}h}\left[d+h-d-{\left({\left(d+h\right)}^2+R^2\right)}^{1/2}+{\left(d^2+R^2\right)}^{1/2}\right]\widehat{i}\\ =\frac{2k_{e}Q}{R^{2}h}\left[h+{\left(d^2+R^2\right)}^{1/2}-{\left({\left(d+h\right)}^2+R^2\right)}^{1/2}\right]\widehat{i}\end{array}$

Conclusion:

Therefore, the electric field created at the same point by a solid cylinder with same dimensions and carrying the same charge is $\frac{2k_{e}Q\widehat{i}}{R^{2}h}\left[h+{\left(d^2+R^2\right)}^{1/2}-{\left({\left(d+h\right)}^2+R^2\right)}^{1/2}\right]$.