Find E Field at Y-Axis: Q1,Q2 Charges at P1,P2

• And 1
In summary, when y= -6.89, the two points E13 and E23 have the same magnitude and opposite direction, and this results in Ex=0.
And 1

Homework Statement

Point charge Q1 = 25nC located at P1(4,-2,7) and a charge Q2 = 60 nC be at P2(-3,4,-2). At what point on the y-axis is Ex=0?

Homework Equations

E= [ Q / (4*pi*E0*r*r) ] ar

The Attempt at a Solution

I took:
( Q1 / 4*pi*E0*(x*x+y*y+z*z) * (y / sqrt(x*x+y*y+z*z)) ay

I got -6.51.

The solution is -6.89.

Am I way off base or on the right path? Thanks for your help.

It appears I might have been way off base. Reworking a different thought...call off the Help Department (for now). :)

Report back in a few...

Ok. I'm stuck again...
Basically we want Ex=0 at some point (P3) on the y-axis. So the point P3 (0,y,0).

Taking the two points to get the total E field. ET = E13 + E23.

I've got:

E13 = 25nC [(0-4)ax + (y+2)ay + (0-7)az ] / (4*pi* 8.854e-12 [(-4)^2 + (y+2)^2 + (-7)^2]^3/2)

E23 = 60nC [(0-(-3)ax + (y-4)ay + (0+2)az ] / (4*pi* 8.854e-12 [(3)^2 + (y-4)^2 + (2)^2]^3/2)

Kind of stuck at this point...any help? Thanks.

Last edited:
I saw throught your work very quickly. Here's what I have to offer:

Now since you want Ex= 0 equate the x components of E13 and E23 together and take their magnitude. Only when they are EQUAL and OPPOSITE in magnitude and direction respectively, will you get Ex= 0.

Yes, that works! You can do it in two ways. From the above post, you can either calculate the value for y, or you can plug in your answer y= -6.89 from the solutions manual that you have and find out that both the magnitudes are the same.

So

25nC* -4
----------------------------- = -0.11927*10^-9 when y= -6.89
[16+ (y+2)^2 + 49 ]^3/2

and

60nC* 3
----------------------------- = -0.11924*10^-9 when y= -6.89
[9+ (y-4)^2 + 4 ]^3/2

I think that's pretty close enough!

I hope that helped.

Yes, it did. Thanks for your help.

1. What is the formula for calculating the electric field at a point on the y-axis?

The formula for calculating the electric field at a point on the y-axis is given by: E = k * Q * y / (y^2 + d^2)^(3/2), where k is the Coulomb's constant, Q is the charge at the point, y is the distance from the point to the y-axis, and d is the distance between the point and the charge.

2. How do you determine the direction of the electric field at a point on the y-axis?

The direction of the electric field at a point on the y-axis is determined by the direction of the force that a positive test charge would experience if placed at that point. If the force would be towards the positive charge, the electric field points towards the positive charge. If the force would be away from the positive charge, the electric field points away from the positive charge.

3. What is the relationship between the distance from the point to the y-axis and the strength of the electric field?

The strength of the electric field at a point on the y-axis is inversely proportional to the square of the distance from the point to the y-axis. This means that as the distance increases, the electric field decreases.

4. How does the presence of multiple charges affect the electric field at a point on the y-axis?

The electric field at a point on the y-axis is the vector sum of the electric fields from each individual charge. This means that the presence of multiple charges can either reinforce or cancel out the electric field at that point, depending on the direction and magnitude of each individual electric field.

5. Can the electric field on the y-axis ever be zero?

Yes, the electric field on the y-axis can be zero if the charges at P1 and P2 are equal in magnitude and opposite in sign, and if the distance between them is equal to the distance between the point on the y-axis and either charge. This is known as an electric field null point.

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