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JEE Main Class 11 Syllabus 2025 PDF for Paper 1 and 2
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Electric Potential Of Uniformly Charged Ring, Rod, And Disc - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Electric potential due to continuous charge distribution(I) is considered one the most difficult concept.

  • 18 Questions around this concept.

Solve by difficulty

QuestionMEDIUM

 A thin disc of radius b=2a has a concentric hole of radius 'a' in it (see figure). It carries uniform surface charge '\sigma 'on it. If the electric field on its axis at height' h ' (h<<a) from its centre is given as 'Ch' then value of 'C' is :

 

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The electric potential at the centre of two concentric half rings of radii $\mathrm{R}_1$ and $\mathrm{R}_{2,}$, having same linear charge density $\lambda$ is :

 

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A spherical conducting shell of radius \mathrm{a}, centred at the origin, has a potential field \mathrm{V=\mathrm{ \begin{cases}V_0 & r \leq a \\ V_0 a / \delta & r<a\end{cases}}} with the zero references at infinity. The stored energy potential to 

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Concepts Covered - 1

Electric potential due to continuous charge distribution(I)

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

\begin{array}{l}{\text { Total charge on ring: } Q} \\ {\text { Charge per unit length: } \lambda=Q / 2 \pi a} \\ \end{array}

Take a small elemental arc of charge dq 

Charge on an arc: dq

So  d V=K \frac{d q}{r}=\frac{Kd q}{\sqrt{x^{2}+a^{2}}}

V(x)=K \int \frac{d q}{\sqrt{x^{2}+a^{2}}}=\frac{K}{\sqrt{x^{2}+a^{2}}} \int d q=\frac{KQ}{\sqrt{x^{2}+a^{2}}}

 

  • The potential at the center of the ring

           V_c=\frac{KQ}{a}  (since x=0)

  •  If x>>a

           V=\frac{KQ}{x}

        As x increases, V will decrease.

         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

 

 

\begin{array}{l}{\text { Total charge on ring: } Q} \\ {\text { Charge per unit Area: } \lambda=Q / \pi R^2} \\ \end{array}

Take a small elemental a ring of radius a having charge as dq 

\begin{array}{l}{\text { Area of ring: } 2 \pi a d a} \\ {\text { Charge on ring: } d q=\sigma(2 \pi a d a)} \\ {\text { Charge on disk: } Q=\sigma\left(\pi R^{2}\right)}\end{array}

\begin{array}{l}{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[\sqrt{x^{2}+a^{2}}]_{0}^{R}=2 \pi \sigma K[\sqrt{x^{2}+R^{2}}-|x|]}\end{array}

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{2KQ}{R}  (since x=0)

  •  If x>>R

         V(x)=2 \pi \sigma K|x|[\sqrt{1+\frac{R^{2}}{x^{2}}}-1] \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{KQ}{|x|}

     As |x| increases, V will decrease.

         As \ \ |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{array}{l}{\lambda=\mathrm{Q} / 2 \mathrm{a} \quad \text { Uniform linear charge density }} \\ {\mathrm{d} \mathrm{Q}=\lambda \mathrm{dy} \text { Charge in length dy }} \\ {\mathrm{dV}=\mathrm{k} \frac{\mathrm{dQ}}{\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{dy}}{\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{1 / 2}}}\end{array}

\begin{aligned} Using \ \ \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] \end{aligned}

We get \mathbf{v}_{\mathrm{P}}=\frac{\mathbf{k} \mathbf{Q}}{\mathbf{2} \mathbf{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] 

 

 

 

 

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Electric potential due to continuous charge distribution(I)

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