# University of Wisconsin Green Bay

Four point charges are fixed in place as shown in the diagram.
a) What is the electric field experienced by the +2 µC charge?
b) What is the force on the +2 µC charge?

• In this problem, you are first asked to find the electric field due to a grouping of point charges. This is a definition problem—the field is caused by the charges and you know an equation that relates them. Quantifying electric field is merely restating the charge and location information that you already have in a slightly different way. Likewise, electric force between point charges is also a definition problem for the same reasons.

Only the 2 µC charge is identified as the system of interest in this problem, because you want to know the field and force on that charge. The other charges are responsible for the interaction—they are the surroundings with which the system interacts.

This problem feels more complicated than many definition problems because the desired quantities are vectors. Therefore, you will need to do the additional math of dividing vectors into components. That math, however, does not change the overall approach to the problem.

• Although many definition problems do not require pictures, you will want them in this case to help you with the vector addition that is required.

Part a):

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Part b):

• Part a)

In equation form, electric field due to a point charge is defined as

E = (kq)/r2

In this case, you will need the vector sum of the field due to three charges.

Part b)

Once you know electric field, the most straightforward way to work part b) is to use the definition of electric force on a charge in an electric field: F = qE.

If you are using this example for a problem in which you are only asked for the force due to point charges, you should use the definition of force due to a point charge

F = (k q1 q2)/r2

and again you will need to find the vector sum of three forces.

Both methods are shown in this example.

• Step 1

Find the electric field due to each point charge from the definition of electric field:
E-3 = (kq-3) / r2
E-3 = (9.0 x 109 N∙m2/C2)( 3 x 10-6 C) / (0.10 m)2
E-3 = 2.7 x 106 N/C

E-2 = (kq-2)/r2
E-2 = (9.0 x 109 N∙m2/C2)( 2 x 10-6 C) / (0.20 m)2
E-2 = 4.5 x 105 N/C

E3 = (kq3)/r2
E3 = (9.0 x 109 N∙m2/C2)( 3 x 10-6 C) / (0.224 m)2
E3 = 5.4 x 105 N/C

Find the x- and y-components of each field in preparation for doing the vector sum:
E3x = E3cos(27o) = -5.4 x 105 N/C cos(27o) = -4.8 x 105 N/C;
E3y = E3sin(27o) = -5.4 x 105 N/C sin(27o) = -2.5 x 105 N/C

E-3x = 0; E-3y = +2.7 x 106 N/C
E-2x = +4.5 x 105 N/C; E-2y = 0

Find the vector sum of the three fields to get the total field:
Ex = E-3x + E-2x + E3x = 0 + 4.5 x 105 N/C + (-4.8 x 105 N/C) = -3 x 104 N/C
Ey = E-3y + E-2y + E3y = 2.7 x 106 N/C + 0 + (-2.5 x 105 N/C) = 2.5 x 106 N/C

E2 = Ex2 + Ey2 = (-3 x 104 N/C)2 + (2.5 x 106 N/C)2 = 6.25 x 1012 N2/C2

E = 2.5 x 106 N/C

tanθ = Ey / Ex = (2.5 x 106 N/C) / (3 x 104 N/C) = 83.3
θ = 89o

The 2 µC charge experiences an electric field that has a magnitude of 2.5 x 106 N/C and is directed at an angle of 89o above the –x axis.

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Step 2

If you have already solved for electric field due to the point charges, as is the case in this problem, the definition relating electric field to electric force is all you need to find force:

The 2 µC charge experiences an electric force that has a magnitude of 5 N and is directed at an angle of 89o above the –x axis.

If you don’t yet have the relation between force and field, you can find force beginning with the point charges.

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Step 3

Find the electric field on the 2µC charge due to each of the other point charges from the definition of electric force:
F2-3 = (kq-3q2) / r2
F2-3 = (9.0 x 109 N∙m2/C2)( 3 x 10-6 C)(2 x 10-6 C) / (0.10 m)2
F2-3 = 5.4 N

F2-2 = (kq-2q2)/r2
F2-2 = (9.0 x 109 N∙m2/C2)( 2 x 10-6 C)(2 x 10-6 C) / (0.20 m)2
F2-2 = 0.9 N

F23 = (kq3)q2/r2
F23 = (9.0 x 109 N∙m2/C2)( 3 x 10-6 C)(2 x 10-6 C) / (0.224 m)2
F23 = 1.1 N

Find the x- and y-components of each field in preparation for doing the vector sum:
F23x = F23cos(27o) = -1.1 N cos(27o) = -0.96 N;
F23y = F23sin(27o) = -1.1 N sin(27o) = -0.49 N

F2-3x = 0; F2-3y = +5.4 N
F2-2x = +0.9 N; F2-2y = 0

Find the vector sum of the three fields to get the total field:
Fx = F2-3x + F2-2x + F23x = 0 + 0.9 N + (-0.06 N
Fy = F2-3y + F2-2y + F23y = 5.4 N + 0 + (-0.49 N) = 4.9 N

F2 = Fx2 + Fy2 = (-0.06 N)2 + (4.9 N)2 = 24 N2

F = 4.9 N

tanθ = Fy / Fx = (4.9 N) / (0.06 N) = 82
θ = 89o

The 2 µC charge experiences an electric force that has a magnitude of 5 N and is directed at an angle of 89o above the –x axis.

• The key to understanding this problem is that both parts are merely definition problems. It feels much more complicated (and takes much more work) because the quantities defined are vector quantities, and you are asked to find the sum of several individual fields (or forces.) Note that almost all of the mathematical effort in this problem is in finding the vector sum. If you don’t remember how to work with vectors and need more of a refresher than is provided in this problem, click here for an example of dividing vectors into components and click here for an example working with the result.

The definition FE = (k q1 q2)/r2 is analogous to Fg = (k m1 m2)/r2. Whenever you are given a distribution of point charges, you can find the force on one of those charges by

1. using the definition to find the magnitude of each force;
2. finding the x- and y-components of each force;
3. adding the forces together separately in each direction.

Electric field (E) is analogous to gravitational field (g). If you know the electric field, you can find the force on any object with a charge q in that field just by the definition FE = qE (analogous to Fg = mg.)

You can also calculate electric field directly from point charges. (If you want to see where the equation comes from, just look at the color coding in the equations above. Remember, red represents the system you care about and blue represents the effect of the surroundings. So electric field contains all information about the surrounding point charges that act on your system.) Whenever you are given a distribution of point charges, you can find the electric field of those charges by

1. using the definition to find the magnitude of each electric field;
2. finding the x- and y-components of each electric field;
3. adding the fields together separately in each direction.