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Electric Potential Due To A Charged Sphere

Electric Potential Due To A Charged Sphere :- We will derive the electric potential due to :

A conducting sphere (solid or hollow) (all charge on the surface)
A uniformly charged non-conducting sphere (with uniform volume charge density)

Each case will be analyzed for :

Points outside the sphere ()
Points on the surface of the sphere (r = R)
Points inside the sphere (

Electric Potential Due To A Conducting Sphere (Solid or Hollow)

(Electric Potential Due To A Charged Sphere)

In a conductor, charges rearrange themselves on the surface such that the electric field inside is zero and the surface is at constant potential. So if the sphere is a conductor, then no matter whether it is hollow or solid, the charge will reside only on its surface. Hence electric potential for a solid conducting sphere and hollow conducting sphere will be same.

Let us assume a conducting sphere of radius R carrying a total charge Q which is uniformly distributed on its surface. The surface charge density (σ) :

$\displaystyle \sigma =\frac{Q}{4\pi {{R}^{2}}}$

(1). Outside the sphere ()

Let us consider a point P outside the sphere at a distance r from it’s center.

We know that relation between electric field intensity and electric potential [Equation (3)] is given by,

$\displaystyle V=-\int\limits_{\infty }^{r}{\overrightarrow{E}.\overrightarrow{dr}}$ …..(1)

For points outside the sphere,

$\displaystyle {{\overrightarrow{E}}_{outside}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{r}^{2}}}\widehat{r}$

$\displaystyle V=-\frac{Q}{4\pi {{\varepsilon }_{0}}}\int\limits_{\infty }^{r}{\frac{1}{{{r}^{2}}}\widehat{r}.\overrightarrow{dr}}$

$\displaystyle V=-\frac{Q}{4\pi {{\varepsilon }_{0}}}\int\limits_{\infty }^{r}{{{r}^{-2}}(1)(dr)\left( cos{{180}^{\circ }} \right)}=\frac{Q}{4\pi {{\varepsilon }_{0}}}\int\limits_{\infty }^{r}{{{r}^{-2}}dr}$

As we move from infinity to point P, we are actually moving in a direction in which r decreases, so we have to replace (dr) by (- dr).

Hence

$\displaystyle V=\frac{Q}{4\pi {{\varepsilon }_{0}}}\int\limits_{\infty }^{r}{{{r}^{-2}}\left( -dr \right)}$

$\displaystyle \Rightarrow V=-\frac{Q}{4\pi {{\varepsilon }_{0}}}\int\limits_{\infty }^{r}{{{r}^{-2}}dr}=-\frac{Q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{r}^{-1}}}{-1} \right]_{\infty }^{r}=\frac{Q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{1}{r}-\frac{1}{\infty } \right]$

$\displaystyle \therefore {{V}_{outside}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{r}$ …..(2)

📌 This is the same as the potential due to a point charge. It means for points outside the sphere, electric potential decreases with distance r and becomes zero at infinity.

(2). At the surface the sphere ()

Putting r = R in equation (1), we get

$\displaystyle {{V}_{surface}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{R}$ …..(3)

(3). Inside the sphere ()

While moving from infinity to a point inside the sphere, the dependence of electric field intensity (E) on distance r is not same. We have to split the integrand given in equation (1) in two parts :

From infinity to the surface (r = ∞ to r = R) &
From the surface to the reference point P (r = R to r = r)

$\displaystyle \therefore V=-\int\limits_{\infty }^{r}{\overrightarrow{E}.\overrightarrow{dr}}=-\left[ \int\limits_{\infty }^{R}{{{\overrightarrow{E}}_{outside}}.\overrightarrow{dr}+}\int\limits_{R}^{r}{{{\overrightarrow{E}}_{inside}}.\overrightarrow{dr}} \right]$

But electric filed inside a charged metallic sphere is zero, i.e., $\displaystyle {{\overrightarrow{E}}_{inside}}=0$ , so

$\displaystyle \therefore V=-\int\limits_{\infty }^{R}{{{\overrightarrow{E}}_{outside}}.\overrightarrow{dr}}$

This will give us the same value as given in equation (3), hence

$\displaystyle {{V}_{inside}}={{V}_{surface}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{R}$ …..(4)

📌 Electric potential inside a conducting sphere (solid or hollow) is same as the value of electric potential on it’s surface and this is the maximum value of electric potential.

Graph between electric potential (V) and distance from the center of the sphere (r) :-

Electric Potential Due To A Uniformly Charged Non-Conducting Sphere

(Electric Potential Due To A Charged Sphere)

Here, the charge is distributed throughout the volume of the sphere uniformly. Let the volume charge density be :-

$\displaystyle \rho =\frac{Q}{\frac{4}{3}\pi {{R}^{3}}}=\frac{3Q}{4\pi {{R}^{3}}}$

The electric field intensity due to a uniformly charged non-conducting sphere at different points is given by :

$\displaystyle {{\overrightarrow{E}}_{outside}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{r}^{2}}}\widehat{r}\text{ }(r>R)$

$\displaystyle {{\overrightarrow{E}}_{surface}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{R}^{2}}}\widehat{r}\text{ }(r=R)$

$\displaystyle {{\overrightarrow{E}}_{inside}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{R}^{3}}}\overrightarrow{r}\text{ }(r<R)$

(1). Outside the sphere ()

$\displaystyle V=-\int\limits_{\infty }^{r}{\overrightarrow{E}.\overrightarrow{dr}}=-\int\limits_{\infty }^{r}{{{\overrightarrow{E}}_{outside}}.\overrightarrow{dr}}=-\frac{Q}{4\pi {{\varepsilon }_{0}}}\int_{\infty }^{r}{\frac{1}{{{r}^{2}}}}\widehat{r}.\overrightarrow{dr}$

As we move from infinity to point P, we are actually moving in a direction in which r decreases, so we have to replace (dr) by (- dr).

Hence

$\displaystyle V=\frac{Q}{4\pi {{\varepsilon }_{0}}}\int\limits_{\infty }^{r}{{{r}^{-2}}\left( -dr \right)}$

$\displaystyle \therefore {{V}_{outside}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{r}$ …..(5)

(2). At the surface the sphere ()

Putting r = R in equation (5), we get

$\displaystyle {{V}_{surface}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{R}$ …..(6)

(3). Inside the sphere ()

Again while moving from infinity to a point inside the sphere, the dependence of electric field intensity (E) on distance r is not same. So we have to split the integrand given in equation (1) in two parts :

From infinity to the surface (r = ∞ to r = R) &
From the surface to the reference point P (r = R to r = r)

$\displaystyle \Rightarrow V=-\left[ \int\limits_{\infty }^{R}{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{r}^{2}}}\widehat{r}.\overrightarrow{dr}+}\int\limits_{R}^{r}{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{R}^{3}}}\overrightarrow{r}.\overrightarrow{dr}} \right]$

$\displaystyle \Rightarrow V=-\left[ \frac{Q}{4\pi {{\varepsilon }_{0}}}\int\limits_{\infty }^{R}{{{r}^{-2}}\widehat{r}.\overrightarrow{dr}+}\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{R}^{3}}}\int\limits_{R}^{r}{\overrightarrow{r}.\overrightarrow{dr}} \right]$

$\displaystyle \Rightarrow V=-\left[ \frac{Q}{4\pi {{\varepsilon }_{0}}}\int\limits_{\infty }^{R}{{{r}^{-2}}(1)(dr)\left( cos{{180}^{\circ }} \right)+}\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{R}^{3}}}\int\limits_{R}^{r}{\left( r \right)(dr)\left( cos{{180}^{\circ }} \right)} \right]$

$\displaystyle \Rightarrow V=-\left[ -\frac{Q}{4\pi {{\varepsilon }_{0}}}\int\limits_{\infty }^{R}{{{r}^{-2}}dr-}\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{R}^{3}}}\int\limits_{R}^{r}{rdr} \right]$

$\displaystyle \Rightarrow V=\left[ \frac{Q}{4\pi {{\varepsilon }_{0}}}\int\limits_{\infty }^{R}{{{r}^{-2}}dr+}\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{R}^{3}}}\int\limits_{R}^{r}{rdr} \right]$

Replacing (dr) by (- dr) because we are integrating in a direction in which r decreases.

$\displaystyle \Rightarrow V=\left[ \frac{Q}{4\pi {{\varepsilon }_{0}}}\int\limits_{\infty }^{R}{{{r}^{-2}}\left( -dr \right)+}\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{R}^{3}}}\int\limits_{R}^{r}{r\left( -dr \right)} \right]$

$\displaystyle \Rightarrow V=-\left[ \frac{Q}{4\pi {{\varepsilon }_{0}}}\int\limits_{\infty }^{R}{{{r}^{-2}}dr+}\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{R}^{3}}}\int\limits_{R}^{r}{rdr} \right]$

$\displaystyle \Rightarrow V=-\left[ \frac{Q}{4\pi {{\varepsilon }_{0}}}\left\{ \frac{{{r}^{-1}}}{-1} \right\}_{\infty }^{R}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{R}^{3}}}\left\{ \frac{{{r}^{2}}}{2} \right\}_{R}^{r} \right]$

$\displaystyle \Rightarrow V=-\left[ -\frac{Q}{4\pi {{\varepsilon }_{0}}}\left\{ \frac{1}{r} \right\}_{\infty }^{R}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{R}^{3}}}\left\{ \frac{{{r}^{2}}}{2} \right\}_{R}^{r} \right]$

$\displaystyle \Rightarrow V=-\left[ -\frac{Q}{4\pi {{\varepsilon }_{0}}}\left\{ \frac{1}{R}-\frac{1}{\infty } \right\}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{R}^{3}}}\left\{ \frac{{{r}^{2}}}{2}-\frac{{{R}^{2}}}{2} \right\} \right]$

$\displaystyle \Rightarrow V=-\frac{Q}{4\pi {{\varepsilon }_{0}}}\left[ -\frac{1}{R}+\frac{1}{{{R}^{3}}}\left\{ \frac{{{r}^{2}}}{2}-\frac{{{R}^{2}}}{2} \right\} \right]$