## Mastering Physics Solutions: Electric Field due to Multiple Point Charges

Electric Field due to Multiple Point Charges

Part A = 0,0.300 N/C
Part B = 0.300 nC

Solutions Below:

Two point charges are placed on the x axis. The first charge, q1 = 8.00 nC, is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.00 nC, is placed a distance 9.00 m from the origin along the negative x axis.

Part A

Calculate the electric field at point A, located at coordinates (0 m, 12.0 m).
Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures.

This is a long problem, but I promise it’s simple if you break it down into parts and solve step by step. Make sure to watch the units since the problem wants the answer in nano Coulombs

The first thing to do is recognize that this is a vector type problem that requires you to set up a triangle in order to solve for the x and y components from each charge. Start by finding the lengths of each side of the triangle.

In the picture below, the side in red (from q2 to point A) has a length of 15m (use the Pythagorean theorem, e.g. sqrt(x2 + y2) = sqrt((-9)2 + 12) and the side in blue has a length of 20m. The base of the triangle is just the distance along the x-axis, which is 25m.

Next, find the angles formed by each side. The angle formed by q1 and the x-axis is 36.87° (sin(θ) = 12m / 20m, so sin-1(12/16) = 36.87°). The angle formed by q2 is 53.13°

Now find the electric field from each charge at Point A, using the formula F = kq / r2. Note that ‘k’ is the Coulomb Constant, 8.9875 * 109 Nm2/C2. Also remember that the charges were given in nC (nano Coulombs, which is to the -9th power):

The electric field from q1 is given by:

F1 = kq1 / r12
F1 = 8.9875 * 109 * (8.00 * 10-9) / 202
F1 = 0.18 N/C

Likewise, for q2:

F2 = kq2 / r22
F2 = 8.9875 * 109 * (6.00 * 10-9) / 152
F2 = 0.24 N/C

To find the x and y components, just use trigonometry (note that the direction of the field is away from the charges, since the charges are positive, so q1‘s x component will be negative, q2‘s x component will be positive, and both y components will be positive:

F1, x = F1 * cos(θ1)
F1, x = 0.18 * -cos(36.87)
F1, x = -0.144 N/C

F1, y = F1 * sin(θ1)
F1, y = 0.18 * sin(36.87)
F1, y = 0.108 N/C

F2, x = F2 * cos(θ2)
F2, x = 0.24 * cos(53.13)
F2, x = 0.144 N/C

F2, y = F2 * sin(θ2)
F2, y = 0.24 * sin(53.13)
F2, y = 0.192 N/C

So the sum of the x components is 0 (+0.144 from q2, -0.144 from q1. The sum of the y components is 0.300 (0.144 + 0.192):

0,0.300 N/C

Part B

An unknown additional charge q3 is now placed at point B, located at coordinates (0 m, 15.0 m).
Find the magnitude and sign of q3 needed to make the total electric field at point A equal to zero.

This is kind of a tricky question because if you just added the total fields from Part A (0.18 and 0.24), you wouldn’t get the right answer. What you have to do is add the x and y components individually, in case they cancel each other out (partially or totally). In Part A you saw that the x-components canceled each other entirely and only the y components remained. Find the overall electric field at Point A by (vector math):

FA = sqrt((Fq1, x + Fq2, x)2 + (Fq1, y + Fq2, y)2)
FA = sqrt((0.144 – 0.144)2 + (0.108 + 0.192)2))
FA = sqrt(0 + 0.09)
FA = 0.3 N/C

Now you can find the charge at Point B needed to zero out the electric field at Point A. Since the electric field at Point A is 0.3 N/C, we need a charge on Point B that will give us this same field value, but in the opposite direction. Since q1 and q2 were both positive, the charge at Point B should also be positive (to push point A back down):

Distance (“r”) between Point B and Point A:

r = 15m – 12m
r = 3m

F = kqB / r2
0.3 = 8.9875 * 109 * qB / 32
0.3 = 8.9875 * 109 * qB / 9
qB = 3.00 * 10-10 C

Since Mastering Physics asks for the answer in nano Coulumbs (-9th power):

0.300 nC

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