# What is the magnitude & direction of the eletric field?

## Homework Statement

Two +3.0nC charges are shown in the diagram below which are spaced 10cm apart. What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ?

What is the electric field in N/C?
What is the direction of the electric field? (above the horizontal)

E = kq/r^2

## The Attempt at a Solution

For the top charge:
E1 = (8.99*10^9)(3*10^-9C)/(.05m)^2 = 10788 N/C
E2 = (8.99*10^9)(3*10-9C)/(.1118m)^2 = 2157.7 N/C

Add them together and I got 1.3 * 10 ^4 N/C

My answer was wrong... was it because the two charge have the same signed charge so really the y-components cancel so the magnitude of the electric field is just 2*E1?

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gneill
Mentor
The electric field is a vector quantity. You need to be adding vectors, not magnitudes.

• sunnnystrong
Doc Al
Mentor
Add them together and I got 1.3 * 10 ^4 N/C
The field contribution from each charge is a vector. You must add them as vectors, not just numbers. Direction counts!

• sunnnystrong
Oh, okay so for the E1 (Top charge):

E1 = ((8.99*10^9)(3*10-9)/(.05^2) i, 0j)

E2 (Bottom charge):
E2 = ((8.99*10^9)(3*10-9)/(.05^2) i, (8.99*10^9)(3*10-9)/(.1^2) j)

Etotal = (21576 i, 2697j)

|Etotal| = 21743.909 N/C ? I already tried this and they said it was wrong

The electric field is a vector quantity. You need to be adding vectors, not magnitudes.
The field contribution from each charge is a vector. You must add them as vectors, not just numbers. Direction counts!

gneill
Mentor
E2 (Bottom charge):
E2 = ((8.99*10^9)(3*10-9)/(.05^2) i, (8.99*10^9)(3*10-9)/(.1^2) j)
This calculation is not correct. You can't take the components of the distance vector and use them individually as distances for the calculation of the field components. You need to use the scalar distance between the charge and the location of interest in the Coulomb equation when finding the magnitude of the field, then break that magnitude into x and y components according to the vector direction. The radial distances are ##r_1## and ##r_2## in the figure below: • sunnnystrong
This calculation is not correct. You can't take the components of the distance vector and use them individually as distances for the calculation of the field components. You need to use the scalar distance between the charge and the location of interest in the Coulomb equation when finding the magnitude of the field, then break that magnitude into x and y components according to the vector direction. The radial distances are ##r_1## and ##r_2## in the figure below:

View attachment 195077
Okay so I am a little confused but for Q1 the electric field would be:
10788 N/C...

for Q2 it would be 2157.73 N/C...

The components of Q1 would be (10788i, 0j) & for Q2 it would be (2157.7cos(60) i, 2157.7sin(60) j) ?

This calculation is not correct. You can't take the components of the distance vector and use them individually as distances for the calculation of the field components. You need to use the scalar distance between the charge and the location of interest in the Coulomb equation when finding the magnitude of the field, then break that magnitude into x and y components according to the vector direction. The radial distances are ##r_1## and ##r_2## in the figure below:

View attachment 195077
& The magnitude of the field would be 1.2*10^4 N/C & the direction would be 8.95 degrees above the horizontal?

gneill
Mentor
Okay so I am a little confused but for Q1 the electric field would be:
10788 N/C...

for Q2 it would be 2157.73 N/C...

The components of Q1 would be (10788i, 0j) & for Q2 it would be (2157.7cos(60) i, 2157.7sin(60) j) ?
Close, but is the angle really exactly 60 degrees? How did you determine the angle?

Note that when you have similar triangles (in the geometric sense of "similar"), it's often easier to just work with the ratios of known side lengths to form the trig functions. So, for example, the cosine of your angle might be formed as:

##cos(θ) = \frac{5}{\sqrt{5^2 + 10^2}}##

• sunnnystrong
Close, but is the angle really exactly 60 degrees? How did you determine the angle?

Note that when you have similar triangles (in the geometric sense of "similar"), it's often easier to just work with the ratios of known side lengths to form the trig functions. So, for example, the cosine of your angle might be formed as:

##cos(θ) = \frac{5}{\sqrt{5^2 + 10^2}}##
ohhh! thank you so much!

the direction of the field will be 9.32 degrees above the horizontal

gneill
Mentor
the direction of the field will be 9.32 degrees above the horizontal
Yes! • sunnnystrong