# Electric field strength at a point due to 3 charges

• Tesla In Person
In summary, the conversation discusses finding the value of a charge that makes the electric field 0, using the equation for electric field. The participant shares their solution, which involves setting the electric field due to the new charge equal to the electric fields of other charges. The result of this calculation is 13q, but the participant expresses uncertainty without the answer to verify. They suggest showing more steps and explaining the choice of signs for the terms in the equation.
Tesla In Person
Homework Statement
Find the value of the charge placed at the point that makes the electric field 0.
Relevant Equations
Electric field: Kq /r^2
I got E. 13q as the answer. That is what i did: The electric field due to +q at origin 0 should equal the electric fields of charges -3q and the new charge placed at 2x. So applying the equation above like this; k*(q) / (2^2) = -3q*k + (k*C)/ 4 solving for C the new charge added, gives 13q. I don't know if it's correct because i don't have the answer to this question . Can you please check my working thanks.

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Tesla In Person said:
Homework Statement:: Find the value of the charge placed at the point that makes the electric field 0.
Relevant Equations:: Electric field: Kq /r^2

I got E. 13q as the answer. That is what i did: The electric field due to +q at origin 0 should equal the electric fields of charges -3q and the new charge placed at 2x. So applying the equation above like this; k*(q) / (2^2) = -3q*k + (k*C)/ 4 solving for C the new charge added, gives 13q. I don't know if it's correct because i don't have the answer to this question . Can you please check my working thanks.

Tesla In Person
In order to make it easier for someone to follow your solution, you could show more steps. In particular, show why x does not appear in your equation and explain the choice of signs for the terms in your equation.

TSny said:
It must be if it really is "Tesla".

Tesla In Person and TSny

## 1. What is the formula for calculating the electric field strength at a point due to 3 charges?

The formula for calculating the electric field strength at a point due to 3 charges is:
E = k * (q1/r1^2 + q2/r2^2 + q3/r3^2)
Where E is the electric field strength, k is the Coulomb's constant, q1, q2, and q3 are the charges of the 3 charges, and r1, r2, and r3 are the distances between the point and each charge.

## 2. How is the direction of the electric field at a point determined when there are 3 charges present?

The direction of the electric field at a point due to 3 charges is determined by the vector sum of the individual electric fields at that point. The direction of the resulting electric field is in the direction of the net force that would be exerted on a positive test charge placed at that point.

## 3. Can the electric field strength at a point due to 3 charges be negative?

Yes, the electric field strength at a point due to 3 charges can be negative. This indicates that the electric field at that point is directed in the opposite direction of the net force that would be exerted on a positive test charge placed at that point.

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

The electric field strength at a point due to 3 charges is inversely proportional to the square of the distance between the point and each charge. This means that as the distance increases, the electric field strength decreases.

## 5. Can the electric field strength at a point due to 3 charges be zero?

Yes, the electric field strength at a point due to 3 charges can be zero. This occurs when the vector sum of the individual electric fields at that point is equal to zero, resulting in no net force on a positive test charge placed at that point.

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