# Calculate Net Electric Field at Point P | Two Charges 42cm Apart

• Himanshu Singh
In summary, the conversation discusses the calculation of the net electric field at point P, which is 15 cm away from a positive charge of 3.2 x 10^-9 C and 27 cm away from a negative charge of -6.4 x 10^-9 C. The correct answer is 1.3 x 10^3 N/C to the right, but the student is confused about the method used to solve it. The summary explains that the electric field is a vector and therefore, the contributions from both charges must be taken into account, considering their direction as well. The student is advised to write out the steps for both paths and to draw arrows on the diagram to visualize the fields. The summary concludes by
Himanshu Singh

## Homework Statement

Two charges, one of 3.2 x 10 ^ -9 C, the other one of -6.4x10^-9 C are 42 cm apart. Calculate the net electric field at point P, 15 cm from the postie charge, on the line connecting the charges

(+) -------------(Point)---------------------- (-)
15cm 27cm

e = kq1/r1^2

## The Attempt at a Solution

I know the correct answer, and the steps involved as well (answer is 1.3 x 10^3 N/C to the right) but I don't get why they do it the way they do it. Can't you just use e = kq1/r1^2 with q being 3.2 x 10 ^ -9 C and d being 0.42m and then subtract that from kq1/r1^2 with q being 3.2 x 10 ^ -9 C and d being 0.27m? I always get the wrong answer with it.

Write out the steps for the two paths in detail and see where you go astray. Post both if you still don't see it.

BvU said:
Write out the steps for the two paths in detail and see where you go astray. Post both if you still don't see it.
Also draw the arrows in the diagram showing the fields that you want to calculate.

Himanshu Singh said:

## The Attempt at a Solution

I know the correct answer, and the steps involved as well (answer is 1.3 x 10^3 N/C to the right) but I don't get why they do it the way they do it. Can't you just use e = kq1/r1^2 with q being 3.2 x 10 ^ -9 C and d being 0.42m and then subtract that from kq1/r1^2 with q being 3.2 x 10 ^ -9 C and d being 0.27m? I always get the wrong answer with it.

Well, the problem is that, the Electric field is a vector and therefore, the net electric field at a point will be the resultant of the electric fields "caused" by both the charges q1 and q2. You are ignoring the other charged particle (q2 = -6.4x10^-9C). As a result you miss out on the contribution due to q2 and get the wrong answer. Account for the both charges and let me know if you get the answer. Do reply if you didn't get this.

Himanshu Singh said:
and then subtract that from kq1/r1^2 with q being 3.2 x 10 ^ -9 C and d being 0.27m?

Since you used the distance from the negative charge, I assume you are referring here to the negative charge and you just typed in the wrong number.

But I think what you're missing is fact that electric field is a vector and you have to take into account the direction of the field.

What direction is the field from a positive charge? What direction is it pointing at point P, right or left?
What direction is the field from a negative charge? What direction is it pointing at point P, right or left?

## 1. What is an electric field?

An electric field is a region in space around a charged object where other charged objects experience a force. It is created by the presence of electric charges and can be either positive or negative.

## 2. How is an electric field measured?

The strength of an electric field is measured in units of Newtons per Coulomb (N/C). This is the force per unit charge that an electric field exerts on a charged object placed in that field.

## 3. What is the relationship between electric field and electric potential?

Electric potential, also known as voltage, is the measure of the electric potential energy per unit charge. Electric potential is directly related to electric field, as the electric field is the gradient of electric potential. In other words, the electric field points in the direction of decreasing electric potential.

## 4. How does the presence of a conductor affect the electric field?

A conductor, such as a metal object, has free electrons that can move easily. When a charged object is placed near a conductor, the electric field causes the free electrons to move, resulting in a redistribution of charges on the surface of the conductor. This creates an electric field inside the conductor that is equal in strength and opposite in direction to the external electric field, effectively canceling it out inside the conductor.

## 5. How does the distance between two charged objects affect the electric field between them?

The strength of the electric field between two charged objects is inversely proportional to the square of the distance between them. This means that as the distance between the objects increases, the electric field decreases. This is known as the inverse-square law.

• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
931
• Introductory Physics Homework Help
Replies
3
Views
857
• Introductory Physics Homework Help
Replies
3
Views
896
• Introductory Physics Homework Help
Replies
68
Views
5K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
18
Views
4K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
2K