# How Do You Calculate Electric Field at Multiple Points Between Two Charges?

• danish1991
In summary: You know the charges themselves. Just plug in the values and add the contributions.In summary, the conversation discusses the calculation of the electric field at point x=1m,2m, and 3m caused by two fixed charges, A and B, located at x axis with values of +4.0uc and +4.0uc respectively. The conversation advises drawing a diagram and calculating the contributions of each charge separately, then summing the results. The relevant equation used is (k)(q)/r, where 'r' is changed to 'r2' to calculate the electric field strength. The directions of the electric fields caused by each charge should be taken into account, and the direction of the field at a given
danish1991

## Homework Statement

Two fixed charges, A and B, are located at x axis. A is at x=0m, B is at x=4m. Qa=+4.0uc and Qb=+4.0uc. Calculate the electric field at point x=1m,2m, 3m.

## Homework Equations

Qa=+4.0x10^-6 Qb=4.0x10^-6
E=F/Q-> E=(k)(q)/r

## The Attempt at a Solution

Can someone help me out with this equation? I haven't been introduced to 3point distances yet. So i don't know how you would set it up, can someone walk me through this? Thanks!

It is always a good idea to draw a diagram to begin with. Pencil in the directions of the electric fields caused by each charge in the places of interest. That will help guide you when it comes time to sum up their contributions at a given point (whether to add or subtract the magnitudes).

Calculate the contributions of each charge separately, one at a time, and then sum the results (this is allowed because electric fields obey the principle of superposition, which you should have learned about as a being a property of linear systems).

Regarding your Relevant Equations, (k)(q)/r gives the electric potential (in Volts) at distance r from a charge q. You're looking for the electric field strength, which has slightly different units: Volts per meter (or equivalently, Newtons per Coulomb). Change the 'r' to 'r2' in your formula.

Okay, but what would i do with the 3 distances mentioned in the last distance, 1m,2m,3m?

gneill said:
It is always a good idea to draw a diagram to begin with. Pencil in the directions of the electric fields caused by each charge in the places of interest. That will help guide you when it comes time to sum up their contributions at a given point (whether to add or subtract the magnitudes).

Calculate the contributions of each charge separately, one at a time, and then sum the results (this is allowed because electric fields obey the principle of superposition, which you should have learned about as a being a property of linear systems).

Regarding your Relevant Equations, (k)(q)/r gives the electric potential (in Volts) at distance r from a charge q. You're looking for the electric field strength, which has slightly different units: Volts per meter (or equivalently, Newtons per Coulomb). Change the 'r' to 'r2' in your formula.

Okay, but what would i do with the 3 distances mentioned in the last distance, 1m,2m,3m?

danish1991 said:
Okay, but what would i do with the 3 distances mentioned in the last distance, 1m,2m,3m?

Take them one at a time and locate them on the diagram that I hope you've drawn. For each of the locations sum the contributions to the electric field from the charges (taking into account their directions, which you penciled in --- right?).

gneill said:
Take them one at a time and locate them on the diagram that I hope you've drawn. For each of the locations sum the contributions to the electric field from the charges (taking into account their directions, which you penciled in --- right?).

Yes, i drawn the diagram. Your going to add them up and the vector is going to be to the right since both charges are positive right?

danish1991 said:
Yes, i drawn the diagram. Your going to add them up and the vector is going to be to the right since both charges are positive right?

That will depend upon where the points are located with respect to the charges. Place vectors on your diagram that indicate the directions of the fields produced by the charges at the points of interest.

The magnitudes of the forces from each charge are easily calculated from the charge and distance from the charge. The direction to assign depends upon the positions of the point and charge. The field "arrows" that you pencil in will tell you the direction.

gneill said:
That will depend upon where the points are located with respect to the charges. Place vectors on your diagram that indicate the directions of the fields produced by the charges at the points of interest.

The magnitudes of the forces from each charge are easily calculated from the charge and distance from the charge. The direction to assign depends upon the positions of the point and charge. The field "arrows" that you pencil in will tell you the direction.

Okay, but how would i set up the equation to the problem? I know the directions of the vectors are going to be facing each other. Qa is going to be toward B, and Qb is going to be towards A. But how would you set up the equation?

danish1991 said:
Okay, but how would i set up the equation to the problem? I know the directions of the vectors are going to be facing each other. Qa is going to be toward B, and Qb is going to be towards A. But how would you set up the equation?

Directions are with respect to the coordinate system that's established. From the given information you know where the X-axis origin is and the position of the charges on the X-axis. So you now know whether the contributions of the fields at any location on the X-axis should be positive or negative (negative if the arrow points to the left, positive if it point to the right).

The rest is just establishing the magnitudes of the field contributions using the formula. You know the positions of the charges and points, so you can establish the distances between the points and charges.

## 1. What is an electric field at a point?

An electric field at a point is a vector quantity that represents the force experienced by a positive test charge placed at that point. It is created by other charges in the vicinity and its strength and direction depend on the magnitude and location of those charges.

## 2. How is the electric field at a point calculated?

The electric field at a point is calculated by dividing the force exerted on a positive test charge by the magnitude of the charge. This results in a vector with the same direction as the force and a magnitude that represents the strength of the electric field at that point.

## 3. What units are used to measure electric field at a point?

The SI unit for electric field is Newtons per Coulomb (N/C). Other common units include Volts per meter (V/m) and Electronvolts per meter (eV/m).

## 4. How does distance affect the electric field at a point?

The strength of an electric field at a point is inversely proportional to the square of the distance from the source charge. This means that as the distance increases, the electric field decreases.

## 5. What is the difference between electric potential and electric field at a point?

Electric potential at a point is a scalar quantity that represents the amount of potential energy per unit charge at that point. Electric field at a point is a vector quantity that represents the force experienced by a test charge placed at that point. In simpler terms, electric potential describes the potential energy of a charge at a point, while electric field describes the force experienced by a charge at that point.

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