Electric Potential Of Uniformly Charged Ring, Rod, And Disc - Practice Questions & MCQ

Electric Potential due to uniformly charged ring

We want to find the electric potential at point P on the axis of the ring as of radius a, shown in the below figure

Total charge on ring: $Q$
Charge per unit length: $\lambda=Q / 2 \pi a$

Take a small elemental arc of charge dq
Charge on an arc: dq

$
\begin{aligned}
& d V=K \frac{d q}{r}=\frac{K d q}{\sqrt{x^2+a^2}} \\
& \text { So } \\
& V(x)=K \int \frac{d q}{\sqrt{x^2+a^2}}=\frac{K}{\sqrt{x^2+a^2}} \int d q=\frac{K Q}{\sqrt{x^2+a^2}}
\end{aligned}
$

- The potential at the center of the ring

$
V_c=\frac{K Q}{a}_{(\text {since } \mathrm{x}=0)}
$

- If $x>>a$

$
V=\frac{K Q}{x}
$

As x increases, V will decrease.

$
\text { As } x \rightarrow \infty, \quad V=0 .
$
 

      So the maximum potential is at the centre of the ring.

 

Electric Potential due to uniformly charged Disc-

We want to find the electric potential at point P on the axis of the disk of radius R, as shown in the below figure

 

 

Total charge on ring: $Q$
Charge per unit Area: $\lambda=Q / \pi R^2$
Take a small elemental a ring of radius a having charge as dq
Area of ring: $2 \pi a d a$
Charge on ring: $d q=\sigma(2 \pi a d a)$
Charge on disk: $Q=\sigma\left(\pi R^2\right)$

$
\begin{aligned}
& d V=K \frac{d q}{\sqrt{x^2+a^2}}=2 \pi \sigma K \frac{a d a}{\sqrt{x^2+a^2}} \\
& V(x)=2 \pi \sigma K \int_0^R \frac{a d a}{\sqrt{x^2+a^2}}=2 \pi \sigma K\left[\sqrt{x^2+a^2}\right]_0^R=2 \pi \sigma K\left[\sqrt{x^2+R^2}-|x|\right]
\end{aligned}
$

We can also write

$
V(x)=\frac{\sigma}{2 \epsilon_0}\left[\sqrt{x^2+R^2}-|x|\right]
$

- The potential at the centre of the disc

$
V_c=\frac{2 K Q}{R}_{\text {(since } \mathrm{x}=0)}
$

- If $x>R$

$
\begin{aligned}
V(x) & =2 \pi \sigma K|x|\left[\sqrt{1+\frac{R^2}{x^2}}-1\right] \simeq 2 \pi \sigma K|x|\left[1+\frac{R^2}{2 x^2}-1\right]=\frac{K \sigma \pi R^2}{|x|} \\
\Rightarrow V(x) & =\frac{K Q}{|x|}
\end{aligned}
$
 

As $|\mathrm{x}|$ increases, V will decrease.

$
\text { As } \quad|x| \rightarrow \infty, \quad V=0
$
 

      So the maximum potential is at the centre of the disc.

Electric Potential due to a finite uniform line of charge-

We want to find the potential due to a finite uniform line of positive charge at point P which is at a distance x from the rod on its perpendicular bisector, as shown in the below figure.

 

$\begin{aligned} & \lambda=\mathrm{Q} / 2 \mathrm{a} \quad \text { Uniform linear charge density } \\ & \mathrm{dQ}=\lambda \mathrm{dy} \text { Charge in length dy } \\ & \mathrm{dV}=\mathrm{k} \frac{\mathrm{d} \mathrm{Q}}{\mathrm{r}} \text { Potential of point charge } \\ & \mathrm{V}_{\mathrm{P}}=\int_{-\mathrm{a}}^{+\mathrm{a}} \mathrm{dV}=\mathrm{k} \lambda \int_{-\mathrm{a}}^{+\mathrm{a}} \frac{\mathrm{dv}}{\left(\mathbf{x}^2+\mathrm{y}^2\right)^{1 / 2}} \\ & \text { Using } \quad \int \frac{d y}{\left(x^2+y^2\right)^{1 / 2}}=\ln [y+r]=\ln \left[y+\left(x^2+y^2\right)^{1 / 2}\right] \\ & \quad \mathbf{v}_{\mathrm{P}}=\frac{\mathbf{k Q}}{\mathbf{2 a}} \ln \left[\frac{\left(\mathbf{x}^2+\mathbf{a}^2\right)^{1 / 2}+\mathbf{a}}{\left(\mathbf{x}^2+\mathbf{a}^2\right)^{1 / 2}-\mathbf{a}}\right]\end{aligned}$