Electric Potential due to Charged Sphere

Electric Potential due to Charged Sphere

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Last updated on August 29th, 2019 at 04:03 pm

The main interest of this article is to calculate the Electric Potential due to Charged Sphere. However, if you are interested in a detailed study of Electric Potential, you can start here. In our previous article, we also have discussed the calculation of Electric Potential due to a Charged Spherical shell. We will use that result to solve this problem. Again, the main idea will be to calculate the amount of work done to move a unit charge from a point of reference to the field of influence of . Our calculation will be very similar to the treatment of Gravitational Potential due to Solid Sphere and we highly recommend to visit that article to have a comparative study.

Link to Gravitational Potential due to Solid Sphere

Electric Potential due to Charged Sphere

Let us consider a uniform solid nonconducting sphere having total charge Q and radius a. The center of the sphere is located at point O. We have to calculate the electric potential due to this charged solid sphere at a test point P, either inside or outside of the sphere. This is shown in the diagram below. We denote the distance between O and P as r. By definition, the potential is the work done to bring a unit charge from infinity to the test point P. The charge density is \rho=\frac{Q}{\frac{4}{3}\pi a^3}

electric potential due to solid sphere at a point outside of the sphere

We will first draw two concentric spherical shells of radius x and x+dx. The volume enclosed by this thick shell of interest is 4\pi x^2dx. Charge enclosed within this thick spherical shell is volume multiplied by the charge density. Thus

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    \[dQ=\frac{Q}{\frac{4}{3}\pi a^3}\times 4\pi x^2 dx\]

    \[=\frac{3Q}{a^3}x^2dx\]

Now we can use the expression we have derived for the Electric Potential due to a Charged Spherical Shell to calculate the Electric Potential due to a Charged Solid Sphere at point P, either outside or inside of the sphere.

Electric Potential at a Point Outside of the Charged Sphere

The most trivial situation is that the test point P is outside of the sphere. We use our ready-made equation from earlier exercise to calculate the potential at this point.

    \[dV=\frac{dQ}{4\pi \epsilon_0 r}\]

Thus the potential due to the solid sphere will be

    \[V=\frac{1}{4\pi \epsilon_0 r}\int{dQ}\]

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

Again we note that the electric potential due to solid sphere at an external point is identical to the electric potential due to a point charge Q situated at the center of the sphere.

Electric Potential at a Point Inside of the Charged Sphere

Here the situation is more complex than the earlier one, however, we can divide the sphere into two parts and use superposition principle to get the final potential. The first sphere has radius r and the second one is a thick spherical shell with an inner radius r and outer radius a. Let us assume that the charge of the inner sphere is Q', then

    \[Q'=\frac{Q}{\frac{4}{3}\pi a^3}\times \frac{4}{3}\pi r^3\]

    \[=\frac{Qr^3}{a^3}\]

electric potential due to solid sphere at a point inside of the sphere

and the potential at a test point, P due to this sphere is

    \[V'=\frac{Q'}{4\pi \epsilon_0 r}\]

    \[=\frac{Qr^2}{4\pi \epsilon_0 a^3}\]

Now we have to calculate the potential at point P due to the spherical shell of the inner radius r and outer radius a. For that, we will divide the shell into several concentric shells. Let us consider one such shell with an inner radius of x and thickness dx. The charge contained in the shell is

    \[dQ''=\frac{Q}{\frac{4}{3}\pi a^3}4\pi x^2 dx=\frac{3Q}{a^3}x^2dx\]

The potential at an internal point P due to this thick shell is

    \[dV''=\frac{dQ''}{4\pi \epsilon_0 x}=\frac{3Q}{4\pi \epsilon_0 a^3}xdx\]

Thus the potential due to this part of the sphere is

    \[V''=\int_{r}^{a}{\frac{3Q}{4\pi\epsilon_0 a^3}xdx}\]

    \[=\frac{3Q}{8\pi \epsilon_0 a^3}(a^2-r^2)\]

and the total electric potential due to the charged solid sphere at an internal point P is

    \[V=V'+V''\]

    \[\boxed{V=\frac{Q}{8\pi \epsilon_0 a^3}(3a^2-r^2)}\]

Electric Potential due to Charged Sphere

In the diagram above, we have plotted electrostatic potential due to a charged sphere as a function of distance from the center of the sphere.

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