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# 02. Analysis Tools

• • Contributed by Paul D'Alessandris
• Professor (Engineering Science and Physics) at Monroe Community College

# Point Charges

Find the electric potential at the indicated point. The charges are separated by a distance 4a.

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The electric potential at the point specified will be the sum of the electric potential from the left charge (VL) and the electric potential from the right charge (VR).

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Notice that since the electric potential is a scalar, calculating the electric potential is often much easier than calculating the electric field.

# Continuous Charge Distribution

The plastic rod of length L at the bottom has uniform charge density l. Find the electric potential at all points to the right of the rod on the x-axis.

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Since electric charge is discrete, the electric potential can always be calculated by summing the electric potential from each of the electrons and protons that make up an object. However, macroscopic objects contain a lot of electrons and protons, so this summation has many, many terms:

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As described earlier, we will replace this summation over a very large number of discrete charges with an integral over a hypothetically continuous distribution of charge. This leads to a relationship for the electric potential at a particular point in space, from a continuous distribution of charge, of:

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where dq is the charge on a infinitesimally small portion of the object, and the integral is over the entire physical object.

The steps for finding the electric potential from a continuous distribution of charge are:

1. Carefully identify and label the location of the differential element on a diagram of the situation.

2. Carefully identify and label the location of the point of interest on a diagram of the situation.

3. Write an expression for dq, the charge on the differential element.

4. Write an expression for r, the distance between the differential element and the point of interest.

5. Insert your expressions into the integral for the electric potential.

6. Carefully choose the limits of integration.

7. Evaluate the integral.

I'll demonstrate below each of these steps for the scenario under investigation.

1. Carefully identify and label the location of the differential element on a diagram of the situation.

The differential element is a small (infinitesimal) piece of the object that we will treat like a point charge. The location of this differential element must be arbitrary, meaning it is not at a "special" location like the top, middle, or bottom of the rod. Its location must be represented by a variable, where this variable is the variable of integration and determines the limits of the integral.

For this example, select the differential element to be located a distance "y" above the center of the rod. The length of this element is "dy". (Later, you will select the limits of integration to go from -L/2 to +L/2 to allow this arbitrary element to "cover" the entire rod.)

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2. Carefully identify and label the location of the point of interest on a diagram of the situation.

You are interested in the electric potential at all points along the x-axis to the right of the rod. Therefore, select an arbitrary location along the x-axis and label it with its location.

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3. Write an expression for dq, the charge on the differential element.

The rod has a uniform charge density l, meaning the amount of charge per unit length along the rod is constant. Since the differntial element has a length dy, the total charge on this element (dq) is the product of the density and the length:

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4. Write an expression for r, the distance between the differential element and the point of interest.

By Pythagoras' theorem, the distance between the differential element and the point of interest is:

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6. Insert your expressions into the integral for the electric potential.

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7. Carefully choose the limits of integration.

The limits of integration are determined by the range over which the differential element must be "moved" to cover the entire object. The location of the element must vary between the bottom of the rod (-L/2) and the top of the rod (+L/2) in order to include every part of the rod. The two ends of the rod form the two limits of integration.

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8. Evaluate the integral.

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This expression looks daunting, but its limiting behaviors (as x approaches zero and x approaches infinity) are correct. Consider the case where x gets smaller and smaller, approaching zero:

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This expression diverges (goes to infinity) as x becomes close to zero, as you should expect since the electric potential directly "on" an electric charge is infinite.

As x gets larger and larger, the L/2 terms become insignificant:

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This expression becomes zero as x becomes infinite, as you should expect since the electric potential extremely far from an electric charge is zero.