Potential Due To Hollow Conductiong, Solid Conducting, Hollow Non Conducting

Electric Potential due to Hollow conducting, Hollow non-conducting, Solid Conducting Sphere-

In the case of Hollow conducting, Hollow non-conducting, Solid conducting Spheres charges always reside on the surface of the sphere.

If the charge on a conducting sphere of radius R is Q. we want to find V at point P at distance r from the centre of the sphere.

• Outside the sphere (P lies outside the sphere. I.e r>R)

$\dpi{100} E_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r^{2}}=\frac{\sigma R^{2}}{\epsilon _{0}r^{2}}$

$V(r)=-\int_{r=\infty}^{r=r} \vec{E}.d \vec{r} =\frac{1}{4 \pi \varepsilon_{0}} \frac{\vec{Q}}{r}$

• Inside the sphere (P lies inside the sphere. I.e r<R )

$E_{in}=0$

$V_{in}=constant$ and it is given as

$\boldsymbol{V}(\boldsymbol{r})=-\int_{r=\infty}^{r=r} \overrightarrow{\boldsymbol{E}} . d \vec{r} =-\int_{\infty}^{R} \boldsymbol{E}_{r}(d \boldsymbol{r})-\int_{R}^{r} \boldsymbol{E}_{r}(\boldsymbol{d} \boldsymbol{r})=\frac{1}{4 \pi \boldsymbol{\varepsilon}_{0}} *\frac{\boldsymbol{q}}{ {\boldsymbol{R}}}+0 \\ \\ \Rightarrow V(r)=\frac{1}{4 \pi \boldsymbol{\varepsilon}_{0}} *\frac{\boldsymbol{q}}{ {\boldsymbol{R}}}$

• At the surface of Sphere (I.e at r=R)

$E_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R^{2}}=\frac{\sigma }{\epsilon _{0}}$

$V_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}=\frac{\sigma R}{\epsilon _{0}}$

• The graph between (E vs r) and (V vs r) is given below

Electric Potential due to Uniformly Charged Non conducting Sphere

Suppose charge Q is uniformly distributed in the volume of a non-conducting sphere of Radius R.

And we want to find V at point P at distance r from the center of the sphere.

• Outside the sphere (P lies outside the sphere. I.e r>R)

$E_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r^{2}}$ $V_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r}$

$E_{out}=\frac{\rho R^{3}}{3 \epsilon _{0}r^{2}}$ $V_{out}=\frac{\rho R^{3}}{3 \epsilon _{0}r}$

• Inside the sphere (P lies inside the sphere. I.e r<R )

$E_{in}=\frac{1}{4\pi \epsilon _{0}}\frac{Qr}{R^{3}}$ $\dpi{100} V_{in}=\frac{Q}{4\pi \epsilon _{0}}*\frac{3R^{2}-r^{2}}{2R^{3}}$

$\dpi{100} E_{in}=\frac{\rho r}{3 \epsilon _{0}}$ $V_{in}=\frac{\rho \left ( 3R^{2}-r^{2} \right )}{6 \epsilon _{0}}$

• At the surface of Sphere (I.e at r=R)

$E_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R^{2}}$ $V_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}$

$\dpi{100} E_{s}=\frac{\rho R}{3 \epsilon _{0}}$ $V_{s}=\frac{\rho R^{2}}{3 \epsilon _{0}}$

Note - If P lies at the centre of the uniformly charged non-conducting sphere (I.e at r=0)

$V_{centre}=\frac{3}{2}\times \frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}=\frac{3}{2}V_{s}$

i.e$\dpi{100} V_{c}> V_{s}$

• The graph between (E vs r) and (V vs r) is given below

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Solved Examples Based On Potential Due To Hollow Conducting, Solid Conducting, Hollow Non-Conducting

Example 1: A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –3Q, the new potential difference between the same two surfaces is

1) V

2)2V

3)4V

4)-2V

Solution:

The electric field in between the shell and sphere is $\\ E \pi x^{2}=\frac{Q_{e n}}{\epsilon_{0}}=\frac{Q}{\epsilon_{0}} \text{(using Gauss's law)} \\ E=\frac{Q}{4 \pi \varepsilon_{0} x^{2}}$

The potential difference between the shells is

$\mathrm{dV}=\mathrm{V}_{\mathrm{r}}-\mathrm{V}_{\mathrm{R}}=\int_{\mathrm{r}}^{\mathrm{R}} \mathrm{Edx}=\int_{r}^{R} \frac{Q}{4 \pi \epsilon_{0} \mathrm{x}^{2}} d x=\frac{Q}{4 \pi \epsilon_{0}}(1 / r-1 / R)$

Thus, $\mathrm{V}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{0}}(1 / \mathrm{r}-1 / \mathrm{R})$

As the potential difference between a solid sphere and a hollow shell depends on the radii of two spheres and the charge on the inner sphere, Since the two values have not changed, the potential difference does not change. Hence the potential difference remains V.

Example 2: A spherical shell of radius R = 1.5 cm has q = 20 $\mu C$ uniformly distributed over it. The force exerted by one half over the other half is

1)0

2)10^-2 N

3)500 N

4)2000 N

Solution:

$\\*E=\frac{\sigma }{2\epsilon_0}\Rightarrow F=E.dQ=\frac{\sigma dA\sigma }{2\epsilon_0}\\*P=\frac{F}{dA}=\frac{\sigma ^{2}}{2\epsilon_0}\\*F=P\pi R^{2}=\frac{\sigma ^{2}}{2\epsilon_0}.\pi R^{2}=\frac{Q^{2}}{32\pi \epsilon_ 0R^{2}}=2000N$

Example 3: A conducting sphere of radius R is given a charge Q. Consider three points B at the surface, A at the centre and C at a distance R/2 from the centre. The electric potential at these points is such that

1) $V_{A}=V_{B}=V_{C}$

2)$V_{A}=V_{B}\neq V_{C}$

3)$V_{A}\neq V_{B}\neq V_{C}$

4)$V_{A}\neq V_{B}= V_{C}$

Solution:

As we learned

Inside the sphere (P lies inside the sphere) -

$E_{in}=0$

$V_{in}=constant$

- wherein

The potential inside a conductor is always constant and equal to the potential at the surface.

Example 4: Which of the following is correct

1)for charged non-conducting sphere $V_{centre}>V_{surface}>V_{out}$

2)for charged non-conducting sphere $V_{centre}\leq V_{surface}>V_{out}$

3)for charged non-conducting sphere $V_{centre}> V_{surface}= V_{out}$

4)for charged non-conducting sphere $V_{centre}> V_{surface}\equiv V_{out}$

Solution:

As we learned

Potential - If P lies at the centre of the uniformly charged non-conducting sphere -

$V_{centre}=\frac{3}{2}\times \frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}=\frac{3}{2}V_{s}$

i.e$\dpi{100} V_{c}> V_{s}$

$V_{centre}=\frac{3}{2}\times \frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Q}{R}=\frac{3}{2}V_{s}$

Example 5: A charge Q is distributed over two concentric conducting thin spherical shells radii r and R (R>r). If the surface charge densities on the two shells are equal, the electric potential at the common centre is ;

1)$\frac{1}{4\pi \epsilon _0}\frac{\left ( 2R+r \right )}{\left ( R^2+r^2 \right )}Q$

2)$\frac{1}{4\pi \epsilon _0}\frac{\left (R+r \right )}{2\left ( R^2+r^2 \right )}Q$

3)$\frac{1}{4\pi \epsilon _0}\frac{\left (R+r \right )}{\left ( R^2+r^2 \right )}Q$
4)$\frac{1}{4\pi \epsilon _0}\frac{\left (R+2r \right )Q}{2\left ( R^2+r^2 \right )}$

Solution:

$\begin{array}{l} \mathrm{Q}_{1}=\sigma 4 \pi \mathrm{r}^{2} \\ \mathrm{Q}_{2}=\sigma 4 \pi \mathrm{R}^{2} \end{array}$

Where ${Q}_{1}$= charge on the inner shell and ${Q}_{2}$= charge on the outer shell

and $Q={Q}_{1}+{Q}_{2}$

So

$\begin{array}{l} Q=\sigma 4 \pi\left(r^{2}+R^{2}\right) \\ \sigma=\frac{Q}{4 \pi\left(r^{2}+R^{2}\right)} \end{array}$

$\begin{array}{l} \mathrm{V}_{\mathrm{C}}=\frac{\mathrm{KQ}_{1}}{\mathrm{r}}+\frac{\mathrm{KQ}_{2}}{\mathrm{R}} =\frac{\mathrm{K} \sigma 4 \pi \mathrm{r}^{2}}{\mathrm{r}}+\frac{\mathrm{K} \sigma 4 \pi \mathrm{R}^{2}}{\mathrm{R}} =\mathrm{K} \sigma 4 \pi(\mathrm{r}+\mathrm{R}) \\ \\ V_C=\frac{\mathrm{KQ} 4 \pi(\mathrm{r}+\mathrm{R})}{4 \pi\left(\mathrm{r}^{2}+\mathrm{R}^{2}\right)} =\frac{\mathrm{KQ}(\mathrm{r}+\mathrm{R})}{\left(\mathrm{r}^{2}+\mathrm{R}^{2}\right)} =\frac{\mathrm{Q}(\mathrm{r}+\mathrm{R})}{4\pi\epsilon _0\left(\mathrm{r}^{2}+\mathrm{R}^{2}\right)}\end{array}$

Summary:
Because material properties differ, and charges are distributed differently in these cases, the electric potential will become different for a hollow conducting, solid conducting, and hollow non-conducting sphere. In the case of a hollow conducting sphere, charges reside on the surface and hence produce a uniform potential inside. For a solid conducting sphere, the charges will distribute on the surface and produce uniform potential inside. In these instances, however, the potential distributions turn out to be different in the case of a hollow non-conducting sphere, for it does not conduct charges. Getting a grip on these differences is very crucial in applications in electronics and physics.