Find the net electric field these charges produce

In summary: Ah, q1 is pointing to the right and q2 to the left, so I subtract the q2 field from the q1 field.One question: Do I include the +/- charges in my equation? ie. would my values for q1 and q2 in the equation be negative since they are both negative charges?I got the write answer by just writing the absolute value of my (negative) answer.edit: completed part (c) with no problems.Yes, you need to include the sign of the charge. In case of a negative charge, the electric field points toward the charge. If it is positive, the electric field points away from it. So the electric field due to q1 points
  • #1
yaro99
75
0

Homework Statement


Two point charges q1 = -6.15 nC, and q2 = -10.5 nC are separated by 25.0 cm (see figure below).
21-p-031-alt.gif


(a) Find the net electric field these charges produce at point A

(b) Find the net electric field these charges produce at point B

(c) What would be the magnitude and direction of the electric force this combination of charges would produce on a proton at A?


Homework Equations


E=ke*Q/r2



The Attempt at a Solution



I got the correct answer for (b), but got an incorrect answer for (a) using the same method.

E = (ke*Q)/(r2)
ƩE = 9*109 * [(6.15*10-9)/(0.152) + (10.5*10-9)/(0.102)] = 11910 N/C to the right
 
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  • #2
yaro99 said:

Homework Statement


Two point charges q1 = -6.15 nC, and q2 = -10.5 nC are separated by 25.0 cm (see figure below).
21-p-031-alt.gif


(a) Find the net electric field these charges produce at point A

(b) Find the net electric field these charges produce at point B

(c) What would be the magnitude and direction of the electric force this combination of charges would produce on a proton at A?


Homework Equations


E=ke*Q/r2



The Attempt at a Solution



I got the correct answer for (b), but got an incorrect answer for (a) using the same method.

E = (ke*Q)/(r2)
ƩE = 9*109 * [(6.15*10-9)/(0.152) + (10.5*10-9)/(0.102)] = 11910 N/C to the right

The electric field is a vector quantity, it has direction. What are the directions of the electric fields due to the individual charges at point (a)?

ehild
 
  • #3
ehild said:
The electric field is a vector quantity, it has direction. What are the directions of the electric fields due to the individual charges at point (a)?

ehild

Ah, q1 is pointing to the right and q2 to the left, so I subtract the q2 field from the q1 field.

One question: Do I include the +/- charges in my equation? ie. would my values for q1 and q2 in the equation be negative since they are both negative charges?
I got the write answer by just writing the absolute value of my (negative) answer.

edit: completed part (c) with no problems.
 
  • #4
Yes, you need to include the sign of the charge. In case of a negative charge, the electric field points toward the charge. If it is positive, the electric field points away from it. So the electric field due to q1 points to the left, that of q2, points to the right. As the field due to q2 is of greater magnitude, the net electric field points to the right, it is positive.

ehild
 
  • #5
(b)

(a) At point A, the net electric field is given by the superposition of the individual electric fields produced by the two charges q1 and q2. We can use the equation E = (ke*Q)/(r2) to calculate the magnitude and direction of this net electric field.

First, we need to find the distance between point A and each of the charges. Using the Pythagorean theorem, we can find that the distance from q1 to point A is 0.152 m and the distance from q2 to point A is 0.102 m.

Plugging these values into the equation, we get:

E = (9*109 * (-6.15*10-9))/(0.152)2 + (9*109 * (-10.5*10-9))/(0.102)2 = -1.85*105 N/C to the left

Therefore, the net electric field at point A is -1.85*105 N/C to the left.

(c) The electric force on a proton at point A would be given by F = qE, where q is the charge of the proton and E is the electric field at point A.

Substituting in the values, we get:

F = (1.6*10-19)(-1.85*105) = -2.96*10-14 N

The force would be directed towards the left, opposite to the direction of the electric field.
 

1. What is a net electric field?

A net electric field is the overall electric field that results from the combination of multiple individual electric fields. It takes into account the magnitude and direction of each individual electric field.

2. How do you find the net electric field produced by multiple charges?

To find the net electric field, you need to calculate the individual electric fields produced by each charge and then add them vectorially. This means taking into account the magnitude and direction of each electric field to determine the overall net electric field.

3. What is the formula for calculating the net electric field?

The formula for calculating the net electric field is: E = k * q / r^2, where E is the net electric field, k is the Coulomb's constant (8.99 x 10^9 N*m^2/C^2), q is the charge of the individual charge, and r is the distance between the charges.

4. Can the net electric field be negative?

Yes, the net electric field can be negative if the individual electric fields produced by the charges are in opposite directions. This means that the overall electric field is pointing in the opposite direction from the majority of the individual electric fields.

5. What is the unit for net electric field?

The unit for net electric field is N/C (newtons per coulomb), which represents the force per unit charge. This unit is the same as the unit for electric field.

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