Question

An insulating sheet carries a charge Q uniformly spread over its surface. The sheet is bent into a perfect half circular arc. A point charge q is located in the center of the arc halfway between the top and bottom of the half cylinder as shown here. The goal is to determine the force of the half cylinder acting upon the point charge.

a) Make a sketch of the problem, establishing a coordinate system and clearly showing the electric field.

b) Sketch the charge element dQ that is used for the integration and show its relationship to the half cylinder.

c) Write E as an integral at the site of charge q.

d) Perform the integration for E and find the force F on point charge q.

e) Find force on q = 3.5 * 10^-7 C assuming the cylinder has dimensions of a piece of paper (8.5 x 11 inches) and has a charge of 6.72 * 10^-5 C. Take the point charge to be 2.5 * 10^-7 C. Is the force stronger when the paper is rolled long and thin or short and fat? (Clearly indicate which calculation is which).

Answer 1

Let us given some parameters to the cylinder

Radius be a units

Height of cylinder be h units

a)

coordinate axis

dQ elemental ring sketch

Electric field due to elemental ring and direction of electric field

b)

Let $\rho$ be the density in unit area

So, for the elemental ring with charge dQ and thickness ds

$\mathrm{d}Q=\rho\pi a\mathrm{d}s$

Now,

$\rho=\frac{Q}{\pi ah}$

$\mathrm{d}Q=\frac{Q}{\pi ah}\pi a\mathrm{d}s$

$\therefore \mathrm{d}Q=\frac{Q}{h}\mathrm{d}s$

c) Let us consider two elemental rings with thickness ds with charge dQ at a distance x from the center as shown in figure

we can see that the forces along other direction get cancel out due to symmetry and only path left is along the negative y axis.

Now, to find the electric field due to the half elemental ring

Let dE' be the field due to point charge dq on the ring of thickness ds' and it is at an angle $\phi$

$\mathrm{d} E'=\frac{1}{4\pi \epsilon _{o}}\frac{\mathrm{d}q}{r^{2}}$

r is the radial distance of the point charge from q.

$r^{2}=a^{2}+x^{2}$

therefore field along the y axis is

$\mathrm{d} E'sin\phi$

Now,

$\mathrm{d} E'sin\phi =\frac{1}{4\pi \epsilon _{o}}\frac{\mathrm{d}q}{r^{2}}sin\phi$

$\Rightarrow \mathrm{d} E'sin\phi =\frac{1}{4\pi \epsilon _{o}}\frac{\mathrm{d}q}{r^{2}}\frac{a}{r}$

$\Rightarrow \mathrm{d} E'sin\phi =\frac{1}{4\pi \epsilon _{o}}\frac{a\mathrm{d}q}{r^{3}}$

Now,

dq is a function of the circumference of the ring.

Let $\lambda$ be the density per unit lenght in the ring

$\mathrm{d}q=\lambda \mathrm{d}s'$

$\Rightarrow \mathrm{d} E'sin\phi =\frac{1}{4\pi \epsilon _{o}}\frac{a\lambda \mathrm{d}s'}{r^{3}}$

$\Rightarrow \oint \mathrm{d} E'sin\phi =\oint \frac{1}{4\pi \epsilon _{o}}\frac{a\lambda \mathrm{d}s'}{r^{3}}$

$\Rightarrow \mathrm{d} E = \frac{1}{4\pi \epsilon _{o}}\frac{a\lambda \left (\pi a \right )}{r^{3}}$

Total charge in the half elemental ring dQ and $\lambda$ is its density

$\mathrm{d}Q=\lambda \pi a$

$\Rightarrow \mathrm{d} E = \frac{1}{4\pi \epsilon _{o}}\frac{a\mathrm{d}Q}{r^{3}}$

d)

$\Rightarrow \oint \mathrm{d} E = \int_{-h/2}^{h/2}\frac{1}{4\pi \epsilon _{o}}\frac{a\mathrm{d}Q}{r^{3}}$

$\Rightarrow E = \int_{-h/2}^{h/2}\frac{1}{4\pi \epsilon _{o}}\frac{aQ\mathrm{d}s}{hr^{3}}$

$\Rightarrow E = \int_{-h/2}^{h/2}\frac{1}{4\pi \epsilon _{o}}\frac{aQ\mathrm{d}s}{h(a^{2}+x^{2})^{3/2}}$

$\Rightarrow E = \frac{2aQ}{4\pi \epsilon _{o}h}\int_{0}^{h/2}\frac{\mathrm{d}s}{(a^{2}+x^{2})^{3/2}}$

$\Rightarrow E = \frac{aQ}{2\pi \epsilon _{o}h}\left | \frac{x}{a (a^{2}+x^{2})^{1/2}} \right |_{0}^{h/2}$

$\Rightarrow E = \frac{Q}{2\pi \epsilon _{o}h}\left | \frac{h/2}{ (a^{2}+(h/2)^{2})^{1/2}} \right |$

$\Rightarrow E = \frac{Q}{\pi \epsilon _{o}}\left ( \frac{1}{ (4a^{2}+h^{2})^{1/2}} \right )$

$\therefore E = \frac{Q}{ \pi \epsilon _{o}(4a^{2}+h^{2})^{1/2}}$