Calculating Electric Field from Two Point Charges

In summary, there are two point charges, one at the origin with a magnitude of 3*10^-6 C and another at a distance of 0.1m on the x-axis with a magnitude of -4*10^06 C. The magnitude of the electric field at a point 0.15m from the origin on the positive y-axis is 88808 N/C. The direction of the electric field can be found using vector addition, which may require finding the coordinates separately or using a vector triangle.
  • #1
infected
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Homework Statement


"A point charge of 3*10^-6 C is at the origin and another point charge of -4*10^06 C is at point 0.1m away along the x axis.
What is the magnitude and direction of the electric field at point 0.15m from the origin along the posting y-axis?


Homework Equations


E= Q/(4PIε°*r^2)
ε°=8.85*10^-12

The Attempt at a Solution


Electric Field from the first charge to the point on the Y axis:
E1 = (3*10^-6)/(4PIε°*0.15^2)

Electric Field from the Second charge to the point on the Y axis:
E2 = (-4*10^-6)/(4PIε°*0.18^2)

Magnitude Of Electric Field = E1+E2=88808N/C

Direction is 158 Degrees from the X(got this from the vectors)

This is my answer, my physics is really rusty and I have a gut feeling I've gone completely wrong. A checking and assistance would be appreciated.

Thanks.
 
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  • #2
welcome to pf!

hi infected! welcome to pf! :smile:

(have a pi: π and try using the X2 and X2 buttons just above the Reply box :wink:
infected said:
"A point charge of 3*10^-6 C is at the origin and another point charge of -4*10^06 C is at point 0.1m away along the x axis.
What is the magnitude and direction of the electric field at point 0.15m from the origin along the posting y-axis?

Electric Field from the first charge to the point on the Y axis:
E1 = (3*10^-6)/(4PIε°*0.15^2)

Electric Field from the Second charge to the point on the Y axis:
E2 = (-4*10^-6)/(4PIε°*0.18^2)

Magnitude Of Electric Field = E1+E2=88808N/C

(you mean the positive y-axis?)

your E1 and E2 magnitudes look ok :smile:

but i think you've added the magnitudes, instead of using vector addition (ie adding the coordinates separately, or using a vector triangle) :wink:
 

1. How do you calculate the electric field between two point charges?

To calculate the electric field between two point charges, you can use the formula E = kq/r^2, where E is the electric field, k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), q is the magnitude of the charges, and r is the distance between the two charges.

2. What is the direction of the electric field between two point charges?

The direction of the electric field between two point charges is determined by the relative positions and charges of the two charges. The electric field points away from positive charges and towards negative charges.

3. Can the electric field between two point charges be negative?

Yes, the electric field between two point charges can be negative. This occurs when the two charges have opposite signs and the electric field points towards the charges instead of away from them.

4. How does the distance between the two point charges affect the electric field?

The electric field between two point charges is inversely proportional to the square of the distance between them. This means that as the distance increases, the electric field decreases.

5. How do you calculate the net electric field from multiple point charges?

To calculate the net electric field from multiple point charges, you can use the principle of superposition. This means that you calculate the electric field from each individual charge and then add them together vectorially to find the total electric field at a specific point.

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