# Electrostatics home work question

• logearav
In summary: Since the net field at a point is the sum of the individual fields, the potential at a point will be zero when the fields from all of the charges at that point add up to zero. For example, if you have two charges at positions A and B, and you want to find the net field at point C, you will take the sum of the fields from both charges at point C, and that will be zero because the fields cancel out. But if you want to find the potential at point C, you will have to take into account the position of your test point relative to both charges at point C, and then calculate the net field accordingly.
logearav

## Homework Statement

Two Charges +q and -3q are separated by a distance of 1 m. At what point in between the charges on its axis is the potential is zero.

## Homework Equations

I formulated the diagram as given in the attachment

## The Attempt at a Solution

At equilibrium V1 = V2
q/4*pi*epsilon*x = -3q/4*pi*epsilon*(r-x)
(r-x) = -3x
x = -r/2 -1/2 = -0.5m. I took r = 1m because distance of separation between the charges is 1 m. Am i correct? How to write the answer because i get negative distance. Could you please help where is the location of point where the potential is zero.

#### Attachments

• diagram.jpg
2.9 KB · Views: 368
You should be finding the value of x where the absolute potentials from the +q and the -3q add to zero, not where these potentials are equal.

mr. misterX, so i should use the equation v1+ v2 = 0
am i right, sir?
which gives me 0.5 m as the answer

That's the right equation to use, but 0.5 m is the wrong answer.

oh sorry! v1= -v2 so i get 0.25 m as the answer. I hope i am right. Then to how to write the answer. Can i write, the potential is zero at a distance .25m from the right of +q.
Some more doubts.
1) can i put -3q on left and +q on the right and solve the above problem. But i don't get 0.25 m
2) Why we use the equation V1 + V2 = 0 at the point O which is a distance x from A?

logearav said:
oh sorry! v1= -v2 so i get 0.25 m as the answer. I hope i am right. Then to how to write the answer. Can i write, the potential is zero at a distance .25m from the right of +q.
Some more doubts.
1) can i put -3q on left and +q on the right and solve the above problem. But i don't get 0.25 m
2) Why we use the equation V1 + V2 = 0 at the point O which is a distance x from A?

1) You have to take into account the relative position on of your "test point" with respect to any given charge in your coordinate system.

Imagine that a point charge is surrounded by field arrows all pointing either radially outward (positive charge) or radially inward (negative charge). The field points IN OPPOSITE DIRECTIONS on opposite sides of the charge. This geometry dependence will necessitate a change of sign in your equation for the potential depending upon your test point location.

It's always best to make a sketch of any given setup of charges, noting the directions of the field contributions of each at places where you want to determine the net field. This will guide you in writing the correct signs for the terms your equations.

2) Electric fields obey the superposition principle. That is, the fields from individual charges can be calculated separately and then summed at a given point to find the net potential at that point.

## 1) What is electrostatics and how does it relate to home work questions?

Electrostatics is the study of stationary electric charges and their interactions. It relates to home work questions because many physics and engineering courses include topics on electrostatics, and students may be required to solve problems and answer questions related to this subject.

## 2) What is the difference between conductors and insulators in electrostatics?

Conductors are materials that allow electric charges to flow easily, while insulators are materials that do not allow electric charges to flow. In electrostatics, conductors are used to transfer and distribute charge, while insulators are used to prevent the flow of charge.

## 3) How do you calculate the electric field in electrostatics?

The electric field is a measure of the force experienced by a charged particle at a given point. It is calculated by dividing the force acting on the charge by the magnitude of the charge. The equation for electric field is E = F/q, where E is the electric field, F is the force, and q is the charge.

## 4) What is the significance of Coulomb's Law in electrostatics?

Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This law is significant in electrostatics because it allows us to calculate the strength of the force between two charged particles, which is crucial in understanding and predicting the behavior of electric charges.

## 5) How does the concept of electric potential relate to electrostatics?

Electric potential is a measure of the amount of work needed to move a charged particle from one point to another in an electric field. In electrostatics, it helps us understand the behavior of electric charges and how they move in an electric field. It also allows us to calculate the potential difference between two points, which is important in determining the direction of charge flow.

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