Electric Potential due to Charged Spherical Shell

Electric Potential due to Charged Spherical Shell

In the main Electrostatic Potential article we have shown the expressions for electric potential due to different charge distribution. In our last article we also have discussed and calculated the Electric Potential due to Point Charge and Charged Ring. Here we will thoroughly discuss the Electric Potential due to Charged Spherical Shell. We will also discuss the situation when the test point is outside of the shell as well as when it is inside of the shell. Notice the very similarities between the treatment and results of electric and gravitational potential. This is because of their inverse square field. You can also browse the Gravitational Potential due to Spherical Shell article to have a comperative study.

An electrostatic potential is the amount of work needed to move a unit positive charge from a point of reference, often chosen as infinity to a test point within the electric field. However we have calculated the electric potential due a charged ring and found it to be  V(r)= \frac{Q}{4\pi \epsilon_0 Z}. We will use this expression to directly calculate the required potential.

Electric Potential due to Charged Spherical Shell

Let us consider a uniform spherical shell of charge Q. The center of this uniform spherical shell is located at point O. We have to calculate the electric potential due this charged spherical system at a test point P, as shown in the figure below. Let us denote the distance between O and P as r.

electric potential due to charged spherical shell

We will first draw a radius a=OA which makes an angle \theta with the line OP. Now we will draw a circle by keeping the central line OA fixed. The radius of this circle (see figure) is aSin\theta. Thus the perimeter of this circle will be 2\pi a Sin\theta. Now consider another circle with radius aSin(\theta+d\theta). These two circles form a ring (shown as green shaded region) having a charge of dQ and width ad\theta. Hence the area subtended by the ring is  2\pi a Sin\theta \times ad\theta = 2\pi a^2 Sin\theta d\theta and the charge is

    \[  dQ=\frac{Q}{4\pi a^2}\times 2\pi a^2 Sin\theta d\theta = \frac{Q}{2}Sin\theta d\theta \]

Considering the triangle OAP, we can write  z^2=a^2+r^2-2ar Cos\theta and taking differential on both side, we find

    \[  Sin\theta d\theta=\frac{zdz}{ar} \]

Thus we can write the expression of dQ in terms of dz as  dQ=\frac{Q}{2ar}zdz.

From our previous exercise, we know that the potential at a point P due to this charged ring is

    \[  dV=\frac{dQ}{4\pi\epsilon_0 z}=\frac{Q}{4\pi\epsilon_0 \times 2ar}dz \]

We can now use this expression to calculate electric potential due to a charged spherical shell at a test point which is either outside or inside of the shell.

Electric Potential at Outside of the Shell

To calculate the potential at a point outside of the shell we have to integrate between \theta=0 to \theta=\pi. This corresponds to <em>z=r-a to <em>z=r+a. Thus,

    \[ V=\frac{Q}{4\pi\epsilon_0 \times 2ar}\int_{r-a}^{r+a}{dz}\\ =\frac{Q}{4\pi\epsilon_0 \times 2ar}[(r+a)-(r-a)] \]

    \[ \boxed{V(\vec{r})=\frac{Q}{4\pi\epsilon_0 r}} \]

This is the final expression of the electric potential due to a charged spherical shell when the test point is outside of the shell. The reult is surprising as the potential due to a charged spherical shell at an external point is identical to the expression of potential due to a point charge Q situated at the center O of the shell.

Electric Potential at Inside of the Shell

As shown in the figure, now we have a situation where \theta=0 corresponds to z=a-r and \theta=\pi corresponds to z=a+r. Thus the potential due to the charged shell at an internal point can be calculated as,

electric potential due to a charged spherical shell at an internal point

    \[ V(\vec{r})=\frac{Q}{4\pi\epsilon_0 \times 2ar}\int_{a-r}^{a+r}{dz}\\ =\frac{Q}{4\pi\epsilon_0 \times 2ar}[(a+r)-(a-r)] \]

    \[ \boxed{V(\vec{r})=\frac{Q}{4\pi\epsilon_0 a}} \]

Surprisingly the electric potential inside a spherical shell is constant and has a value of \frac{Q}{4\pi\epsilon_0 a}.

 

electric potential due to charged spherical shell

In the diagram above we have shown the electrostatic potential due to a spherical shell as a function of distance from the center of the shell.

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