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Recall from last lecture:

Example P15.4

An alpha particle (charge = +2.0e) is sent at high speed toward a gold nucleus (charge = +79e). What is the electrical force acting on the alpha particle when it is 2.0×10-14m from the gold nucleus?

First, sketch an alpha particle moving towards a gold nucleus. For the purposes of this course, electrons, protons, and nuclei can be considered to be point charges. An alpha particle is a helium nucleus.

We'll use Coulomb's law to calculate the electrical force.
F = kq1q2/r² = (9×109Nm²/C²)(2.0e)(79e)/(2.0×10-14m)² = (360×1037N/c²)(1.6×10-19C)2 = 91N. This force is repulsive since both charges (alpha particle and gold nucleus) are positive.

The Superposition Principle

When dealing with 3 or more charges, the net force on one of them is the vector sum of the forces due to each of the others taken individually. For example, if we have charges called 1, 2, 3, ..., the total force exerted on charge 1 is:

Ftot = F12 + F13 + ...
F12 is the force on charge 1 from charge 2. This is a vector sum; you must sum the x and y components of the forces separately.

Example P15.8

A 2.2×10-9C charge is on the x axis at x = -1.5m and a 5.4×10-9C charge is on the x axis at x = 2.0m. Find the net force exerted on a 3.5×10-9C charge located at the origin.

Start by making a sketch of the problem. Draw the x axis and the locations of the 3 charges. Show the forces exerted on the charge at the origin; there are 2 of them, and since all the charges are positive, all the forces are repulsive. One force points left and the other points right.

F1 = kq1Q/r²1 = (9.0×109Nm²/C²)(2.2×10-9C)(3.5×10-9C)/(1.5m)² = 31×10-9N = 3.1×10-8N. F1 points to the right (+x direction).

F2 = kq2Q/r²2 = (9.0×109Nm²/C²)(5.4×10-9C)(3.5×10-9C)/(2.0m)² = 43×10-9N = 4.3×10-8N. F2 points to the left (-x direction).

The net force is Ftot = F1 - F2 = -1.2×10-8N. F2 is subtracted because it points to the left, in the -x direction.

16.4 The Electric Field

The electric field is something that upon first sight seems like little more than slight of hand. We define the electric field at a point in space to be the force that would be produced on a test charge at that location, divided by the value of the test charge:

E = |F| / |q0|
where q0 is the test charge, and F is the force on it.

So what have we gained? At first sight, very little. However, as we proceed through the study of electricity and magnetism, the electric field takes on a reality of its own, until joining with the magnetic field in electromagnetic waves.

Since the force on a test charge q0 from a charge q is F = |q| }q0| / r², the electric field due to q is:

E = k |q| / r².
The electric field is a vector quantity, and, by convention, it points in the same direction as would the force produced on a positive test charge. The electric field has units of N/C.

The electric field obeys the superposition principle exactly as do forces. Therefore, the net electric field due to two or more charges is given by summing the electric field produced by each charge individually. Again, you must sum the x and y components because E is a vector quantity.

Example: P15.8 (modified)

A 2.2×10-9C charge is on the x axis at x = -1.5m and a 5.4×10-9C charge is on the x axis at x = 2.0m. Find the electric field at the origin.

Label the charge at -1.5m number 1, and the charge at 2.0m number 2. Determine the electric field due to each, and sum it up.

E1 = k q1 / r1² = (9.0×109 Nm²/C²)(2.2×10-9C)/(1.5m)² = 8.8 N/C. E1 points in the +x direction.

E2 = k q2 / r2² = (9.0×109 Nm²/C²)(5.4×10-9C)/(2.0m)² = 12.2 N/C. E2 points in the -x direction.

Etot = E1 - E2 = 8.8 N/C - 12.2 N/C = -3.4 N/C.

What is the force on a 3.5×10-9C charge at the origin?

F = qE = (3.5×10-9C)(-3.4 N/C) = -12×10-9 N. This is the same value we calculated earlier, as it must be.

Copyright © Robert Harr 2013