Mastering physics: The Electric Field at a Point Due to Two Point Charges

In summary, the problem involved calculating the net electric field at the origin due to two point charges, q1=-4.00 nC and q2=+6.00 nC, located at specific points. The magnitude of the net electric field was calculated using the equation E=K|q|/r^2, with K=8.988x10^9 and r=sqrt(0.600^2 + 0.800^2), resulting in a value of 35.95 N/C. The direction of the electric field was also asked for, relative to the negative x-axis, and was expressed in degrees. This part was not attempted by the individual.
  • #1
silver421
4
0

Homework Statement


A point charge q1:-4.00 nC is at the point x=0.600 meters, y=0.800 meters, and a second point charge q2:+6.00 nC is at the point x=600 meters, y=0 .

a) Calculate the magnitude of the net electric field at the origin due to these two point charges.
Express your answer in Newtons per coulomb to three significant figures.

b) What is the direction, relative to the negative x axis, of the net electric field at the origin due to these two point charges.
Express your answer in degrees to three significant figures.

[tex]\phi[/tex]= _________ (up from the negative x axis)


Homework Equations





The Attempt at a Solution


I tried calculating E1 and E2 and added them both to get E but my answers are wrong
 
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  • #2
That is the correct way, as long as you take into account the fact that the electric field has a direction as well.

Can you show us how you calculated, for example, E1?
 
  • #3
i used the equation E=K|q|/r^2. i used these numbers K= 8.988x10^9 q= 4.00x 10^-9 nC
r= square root of: (0.600)^2 + (0.800)^2
and i got E1= 35.95

and for the angle part i have no idea how to do it!
 

1. What is the electric field at a point due to two point charges?

The electric field at a point due to two point charges is the force per unit charge experienced by a test charge placed at that point. It is calculated by taking the sum of the electric fields produced by each individual point charge, taking into account the distance and direction from the charges to the point in question.

2. How do the magnitudes and signs of the point charges affect the electric field at a point?

The magnitudes and signs of the point charges have a direct impact on the electric field at a point. The greater the magnitude of the point charges, the stronger the electric field will be. Additionally, the sign of the charges will determine the direction of the electric field. Like charges will repel each other and create an outward electric field, while opposite charges will attract each other and create an inward electric field.

3. What is the equation for calculating the electric field at a point due to two point charges?

The equation for calculating the electric field at a point due to two point charges is E = k * (q1/r1^2 + q2/r2^2), where k is the Coulomb's constant, q1 and q2 are the magnitudes of the point charges, and r1 and r2 are the distances from the point charges to the point in question.

4. Can the electric field at a point due to two point charges be negative?

Yes, the electric field at a point due to two point charges can be negative. This occurs when the magnitude of the electric field produced by one point charge is greater than the other, resulting in a net electric field in the opposite direction. This can also occur when the two point charges have opposite signs and the resulting electric field is directed towards the charges rather than away from them.

5. How does the distance between the point charges affect the electric field at a point?

The distance between the point charges has a significant impact on the electric field at a point. As the distance between the charges increases, the electric field becomes weaker. This is because the electric field follows an inverse-square law, meaning that the strength of the field decreases exponentially as the distance increases. Therefore, the closer the point charges are to each other, the stronger the electric field will be at a point located between them.

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