Question

Potential of a Charged Cylinder Part A A hollow cylinder of radius r and height h has a total charge q uniformly distributed over its surface. The axis of the cylinder coincides with the z axis, and the cylinder is centered at the origin, as shown in the figure. (Figure 1) What is the electric potential V at the origin? View Available Hint(s) Figure 1 of 1 > V= In 2 coh 41% 2r h + V = 4 In( h 27 1+ ha 42 2πεoh V 4r? 9 In 2περή 21 h + 1+ V = rent In (+ 1+ ha 42 2περή V . Submit Previous Answers Correct

Need help in PART B. kindly write the solution as well. Thanks

Answer 1

Po d /d²+r2 r o 2. dr

Let us calculate electric potential at P due to a charged ring of radius r. Point P is at a distance d from center of ring.

For a small charge element $\lambda$ dr , potential dV at P is given by

AP = K Adr 72 +d

where $\lambda$ is linear charge density , i.e., charge density per unit length and

K = 1/(4) is Coulomb's constant

TEO

Electric potential V due to full ring is given as

= | av = dV = KX dr Vr? + d2 27 KX Vr? + d2

since $\lambda$ $\times$ (2$\pi$r) is the charge q in the ring , potential V due to ring at a axial distance d is written as

...........................(1)

V = K 9 Vr2 + d2

----------------------------------------------------

Now let us find the electric potential of a charged hollow cylinder at mid point of its axis. Let r be the radius of hollow cylinder and h is its height.

Let us fix origin of coordinate system at mid point of its axis as shown in figure .

ds S Y >X h

Let us consider a ring of thickness ds at a distance s from origin

electric potential dV due to this ring is written using eqn.(1) as

$dV = K \frac{2\pi r\sigma ds}{\sqrt{s^{2}+r^{2}}}$

where $\sigma$ is surface charge density

Potential due to full hallow cylinder is given by

$V = \int dV = K (2\pi r\sigma ) \int_{-h/2}^{h/2}\frac{ds}{\sqrt{s^{2}+r^{2}}} = \frac{q}{2\pi \epsilon _{o}h} log\left ( \frac{h}{2r} + \sqrt{1 + \left ( \frac{h^{2}}{4r^{2}}\right )} \right )$

Above integration is performed using substitution s = r tan$\theta$ and we used the dfinition of surface charge density

$\sigma = \frac{q}{2\pi rh}$

--------------------------------------------------------------

When h $\rightarrow$ 0 , hallow cylinder becomes ring .

Let us consider the potential function

$V = \frac{q}{2\pi \epsilon _{o}h} log\left ( \frac{h}{2r} + \sqrt{1 + \left ( \frac{h^{2}}{4r^{2}}\right )} \right )$.........................(2)

when h $\rightarrow$ 0 ,

$\sqrt{1 + \left ( \frac{h^{2}}{4r^{2}}\right )} = \left ( 1 + \left ( \frac{h^{2}}{4r^{2}}\right ) \right )^{1/2} = 1 + \frac{1}{2}\left ( \frac{h^{2}}{4r^{2}}\right )$

Hence eqn.(2) is written as

$V = \frac{q}{2\pi \epsilon _{o}h} log\left ( 1+ \frac{h}{2r} + \frac{1}{2}\left ( \frac{h}{2r}\right )^{2} \right )$ ...........................(3)

In above equation , argument of log function is $e^{h/2r}$ by neglecting higher powers of h

Hence eqn.(3) will become

$V = \frac{q}{2\pi \epsilon _{o}h} \times \frac{h}{2r} = \frac{q}{4\pi \epsilon _{o}r}$

Above equation for potential is same as potential of ring at centre