Electric Potential - Conceptual question

[EngineerHead]

Homework Statement

At a certain distance from a point charge, the magnitude of the electric field is 500 V/m and the electric potential is -3.00 kV. (a) What is the distance to the charge? (b) What is the magnitude of the charge?

Homework Equations

The Attempt at a Solution

a) 6 meters
b) -2 micro-Coulombs
*These are correct, I mention this fact cause I don't want you to have to waste your time checking so.

My question is: for part b, if you solve for Q, the magnitude of charge, from E instead of V (E = kq/r^2), what is the visual concept behind why the charge is negative? Because obviously if you solve from V, you will get a negative 2 uC.

Of course the book asks for magnitude anyway, but the answer key gives the negative sign as well. I want to understand the concept behind it.

 

[Doc Al]

My question is: for part b, if you solve for Q, the magnitude of charge, from E instead of V (E = kq/r^2), what is the visual concept behind why the charge is negative?

You cannot tell the sign of Q just using E, since you were not told the direction of the field.

 

[EngineerHead]

Thank you, and just to verify - because of the potential given, it is a fact that the charge is negative?

 

[Doc Al]

Thank you, and just to verify - because of the potential given, it is a fact that the charge is negative?

Yes.

 

[asteriatic]

The concept behind the negative sign in the magnitude of charge is related to the direction of the electric field. The electric field points in the direction of the force that a positive test charge would experience if placed in the field. In this case, the electric field points away from the point charge, indicating that a positive test charge would experience a repulsive force. This means that the point charge itself must have a charge of the opposite sign, in this case negative, in order to create the electric field. This is why the magnitude of the charge is negative when solving for it using the electric field equation.