Electric Field due to Two Point Charges

In summary, two point charges are placed on the x axis, q1=8 nC at 16m from the origin along the positive x axis, and q2=6 nC at 9m from the origin along the negative x axis. The electric field at the origin, point O, is found to be 0.928 N/C, with an x component of 0.667 N/C and a y component of -0.261 N/C. When q2 is changed to -6 nC, the net electric field at the origin is 0.406 N/C, with an x component of 0.667 N/C and a y component of -0.667 N/C. The direction of the
  • #1
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Homework Statement



Two point charges are placed on the x axis. The first charge, q1= 8 nC, is placed a distance 16m from the origin along the positive x axis; the second charge, q2= 6 nC , is placed a distance 9m from the origin along the negative x axis.

Find the electric field at the origin, point O.

Give the x and y components of the electric field as an ordered pair. Express your answer in Newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.

Now, assume that charge q2 is negative; q2 = -6 nC. What is the net electric field at the origin, point O?

Give the x and y components of the electric field as an ordered pair. Express your answer in Newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.

Homework Equations



E = [tex]\frac{kQ}{r^2}[/tex]

The Attempt at a Solution



I start by "placing" a test charge at the origin.

Here's what I think I should do:

E = [tex]\frac{kQ_1}{r^2}[/tex] + [tex]\frac{kQ_2}{r^2}[/tex]

E_1 = [tex]\frac{(9x10^9)(8x10^-9)}{16^2}[/tex] = 0.261 N/C

E_2 = [tex]\frac{(9x10^9)(6x10^-9)}{9^2}[/tex] = 0.667 N/C

E = E_1 + E_2 = 0.261 + 0.667 = 0.928 N/C

Am I right?
 
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  • #2
The direction of the E-field is taken to be the direction of the force it would exert on a positive test charge so when you draw a diagram with a positive test charge how would the forces from the q1 and q2 be acting on it? In the same direction or opposite?
 
  • #3
Midy1420 said:
The direction of the E-field is taken to be the direction of the force it would exert on a positive test charge so when you draw a diagram with a positive test charge how would the forces from the q1 and q2 be acting on it? In the same direction or opposite?

Since all of the charges are positive, they would repel, and thus create a force pushing it to the left.

Still not sure (electircity isn't my strong point)
 
  • #4
q1 is to the right of the test charge (placed at the origin) and q2 is to the left of it.. draw a force diagram to see
 
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  • #5
ohhhh... I thought the second charge was also placed on the positive x-axis

so would it be 0.667 - 0.261 since the force is pushing the test charge to the left?

and I just noticed that I posted it in the wrong section. Would someone be able to move it to intro physics?
 
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What is the formula for electric field due to two point charges?

The formula for electric field due to two point charges is given by E = k*q/(r^2), where k is the Coulomb's constant, q is the charge of one of the point charges, and r is the distance between the two point charges.

How do you calculate the direction of electric field due to two point charges?

The direction of electric field due to two point charges is given by the direction of the force experienced by a positive test charge placed in the electric field. The direction of force can be determined using the right-hand rule, where the fingers point in the direction of the test charge's velocity and the thumb points in the direction of the force.

What happens to the electric field if the two point charges have the same sign?

If the two point charges have the same sign (both positive or both negative), the electric field will be repulsive. This means that a positive test charge placed in the electric field will experience a repulsive force, and a negative test charge will experience an attractive force.

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

The electric field is inversely proportional to the square of the distance between two point charges. This means that as the distance between the point charges increases, the electric field decreases. This is because the electric field spreads out over a larger area as the distance increases.

Can the electric field be zero between two point charges?

Yes, the electric field can be zero between two point charges if the two point charges have equal magnitude but opposite signs. In this case, the electric fields created by each point charge cancel each other out, resulting in a net electric field of zero in between the two charges.

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