# Magnitude & Direction of electric fields?

• daimoku
In summary, the question is asking for the magnitude and direction of the electric field at the center of a square with a given charge and distance. Using the Pythagorean theorem, the distance to the center is calculated and then the electric field is calculated for each corner of the square. To find the net electric field, the x and y components are summed, resulting in a magnitude of 87984.636 N/C. However, since there is no y-component, the angle is 0 degrees counterclockwise from the horizontal.
daimoku

## Homework Statement

What are the magnitude and direction (in degrees counterclockwise from the horizontal) of the electric field at the center of the square of Fig. 23-31 if q = 0.9*10^-8 C and a = 5.1 cm?

http://personalpages.tds.net/~locowise/23-30.gif

## Homework Equations

E=kq/r^2
k=8.99*10^9 N*m^2/C^2

## The Attempt at a Solution

Calculated r using the pythagorean theorem:
5.1cm--->.051m
sqrt(.051^2+.051^2)=.07212m ...this is the length of the entire diagonal so r would be the diagonal divided by 2 which equals .036062445841 meters.

Starting with E1 (the upper left hand corner of the square) and moving clockwise to E4...
E1=+62214.53 N/C
E2=-124429.07 N/C
E3=+124429.07 N/C
E4=-62214.53 N/C

Here is where I get confused...do I resolve these in horizontal and vertical components? And, how do I find the angle? Thanks for your help!

Last edited by a moderator:
I believe you would do well to define an xy-coordinaate system, with origin at the center of the square. The electric field is a vector, and thus will have x and y components. You want to find all of the x-components of the field at the center of the square, and sum them, and do the same for the y-components. Once you have these two components, you can use the Pythagorean theorem to find the magnitude of the electric field. Also, at this point, it will be easy to find the angle at which the field vector points (because you will know which directions the components point), and you can use the arc-tangent to find this. Do you follow?

Thanks for your tips. Here's what I got:
Xnet= -87984.636 N/C
Ynet= 0 N/C
Using the pythagorean theorem:
Enet= 87984.636 N/C
This much is correct but when I take the arctangent(0/-87984.636) I get 0. Any ideas?

Last edited:
If the x net is -87984.636, and there is zero net in the y-direction, then you simply have a vector pointing in the negative x-direction with the x-net magnitude. There will be no y-component and thus the vector cannot rise above the x-axis, and so the angle it makes with the horizontal is zero, right?

That makes sense but I must have made a mistake somewhere along the way. The magnitude is correct but the angle isn't. Here's my math:

+q E1=62214.53 N/C
-2q E2=-124429.0657 N/C
-q E3=-62214.53 N/C
+2q E4=124429.0657 N/C

E1
X=62214.53sin135=43992.318 N/C
Y=62214.53cos135=-43992.318 N/C
E2
X=-124429.0657sin45=-87984.636 N/C
Y=-124429.0657cos45=-87984.636 N/C
E3
X=-62214.53287sin225=43992.318 N/C
Y=-62214.53287cos225=43992.318 N/C
E4
X=124429.0657sin315=-87987.636 N/C
Y=124429.0657cos315=87987.636 N/C

Xnet and Ynet are the values listed above. Maybe I overlooked something? Thanks for your quick responses! The angle is 90 degrees above the horizontal...I don't really understand why though.

Last edited:

## 1. What is the meaning of magnitude in electric fields?

The magnitude of an electric field refers to the strength or intensity of the electric field at a certain point in space. It is measured in units of newtons per coulomb (N/C) and represents the force experienced by a unit charge placed in the field.

## 2. How is the direction of an electric field determined?

The direction of an electric field is determined by the direction that a positive test charge would experience a force when placed in the field. The direction is always perpendicular to the equipotential lines and points from positive to negative charges.

## 3. Can the direction of an electric field change?

Yes, the direction of an electric field can change depending on the location of the charges creating the field. It is always directed away from positive charges and towards negative charges, but the specific direction may change depending on the arrangement of charges.

## 4. How does the distance from a charge affect the magnitude of the electric field?

The magnitude of an electric field is inversely proportional to the square of the distance from the charge. This means that as the distance increases, the electric field decreases. This relationship is known as the inverse square law.

## 5. What factors can affect the direction and magnitude of an electric field?

The direction and magnitude of an electric field can be affected by the distance from the source charge, the strength of the source charge, and the presence of other charges in the vicinity. Additionally, the presence of conductors or insulators in the surrounding area can also affect the electric field.

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