Finding the magnitude of the electric field from 3 charges at a point P

• KKuff
In summary, the problem involves finding the magnitude of the electric field at point P, which is 80 nm from Q2 on a line perpendicular to the line connecting Q1 and Q3. Using the given charges and distances, the correct answer is 1.9 x 106 N/C. To arrive at this answer, the poster calculated the magnitude of the electric field for each charge and then found the resultant vector using trigonometry. However, the incorrect angle was used in the calculation of the y-component of Q1 and Q3, resulting in an incorrect answer. The correct angle to use is 80/100, which gives a correct answer of 1.9 x 106 N/C. In the future, the poster
KKuff

Homework Statement

Hi, here is the problem I'm having trouble with:
Three charges Q1, Q2, and Q3, each equal to 6.4 × 10–19 C, are in a straight line. The distance between neighboring charges is 60 nm. Find the magnitude of the electric field at P, which is 80 nm from Q2 on a line at right angles to the line between Q1 and Q3

E = kQ/r2

The Attempt at a Solution

The correct answer is 1.9 x 106 N/C, but I cannot figure out how to arrive at that answer.
I started by first finding the magnitude of the electric field from each charge
E1 = kQ1/r2 = (8.99 x 109 Nm2/C2)(6.4 x 10-19C) / (100 x 10-9m)2 = 575360 N/C

E2 = kQ2/r2 = (8.99 x 109 Nm2/C2)(6.4 x 10-19C) / (80 x 10-9m)2 = 899000 N/C

E3 = kQ3/r2 = (8.99 x 109 Nm2/C2)(6.4 x 10-19C) / (100 x 10-9m)2 = 575360 N/C

Then I tried splitting those into the x and y components
Ex = 0, since E1x and E3x cancel each other out and there is no x component of E2

E1y = 575360sin$\theta$ = 345216 N/C
E3y would also equal this
E2y = 899000 N/C

Ey = 1589432

so to find the magnitude of the resultant vector...
E = $\sqrt{}$(Ex2 + Ey2)
E = $\sqrt{}$(02 + 15894322) = 1589432
E = 1.59 x 106

So what am I doing wrong here? Any help would be appreciated

The magnitudes of the fields for the three charges look okay, but the values you've calculated for the y-components of charges Q1 and Q3 don't look right. Check the angle you've used in the sine function.

Hi, thanks for the reply, I got sin$\theta$ = 0.6, using the dimensions of the right triangle with the distance between Q1 and Q2 being 60nm and the distance between Q2 and P being 80nm, so the hypotenuse would be 100nm 602 + 802 = 1002. Then sin$\theta$ = 60nm / 100nm. Is this wrong?
Thanks

sin(θ) = 80/100

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Thanks, I was using the wrong angle. In the future, how should I go about choosing the correct angle?

KKuff said:
Thanks, I was using the wrong angle. In the future, how should I go about choosing the correct angle?

You can use either angle as long as you use the correct trig function! The trig functions are just ratios, as you know.

When you've got all the side lengths for the triangle (as you do in the case of this problem) it's probably easier to just form the correct ratio directly from it; Choose the side that lies in the same direction as the vector component you're looking for and divide by the hypotenuse.

1. How do you calculate the magnitude of the electric field from 3 charges at a point P?

To calculate the magnitude of the electric field at a point P, we use the formula E = kQ/r^2, where k is the Coulomb constant, Q is the charge of the source, and r is the distance between the source and point P. We calculate the electric field from each of the 3 charges and then add them together using vector addition to find the total magnitude at point P.

2. What is the Coulomb constant and how is it used in this calculation?

The Coulomb constant, denoted by k, is a proportionality constant used in the calculation of the electric field. Its value is approximately 8.99 x 10^9 Nm^2/C^2. It is used to convert the units of charge and distance into the correct units for the electric field calculation.

3. Can the magnitude of the electric field be negative?

Yes, the magnitude of the electric field can be negative. This indicates that the direction of the electric field is opposite to the direction of the electric force on a positive test charge. The magnitude of the electric field is always positive, but its direction can be negative.

4. How does the distance between the charges and point P affect the magnitude of the electric field?

The magnitude of the electric field is inversely proportional to the square of the distance between the charges and point P. This means that as the distance increases, the magnitude of the electric field decreases. This relationship is described by the inverse square law.

5. Is the magnitude of the electric field affected by the charges of the particles?

Yes, the magnitude of the electric field is directly proportional to the charges of the particles. This means that as the charge of a particle increases, the magnitude of the electric field also increases. This relationship is described by Coulomb's law.

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