What is the electric field magnitude at points where electric potential is zero?

[n_h0987]

Homework Statement

A -14.3 nC point charge and a +24.9 nC point charge are 17.4 cm apart on the x-axis. What is the magnitude of the electric field at the two points on the x-axis where the electric potential is zero?

Homework Equations

V = kq/r
E = kq/r^2

The Attempt at a Solution

Since the potential = 0,

q1 = -14.3 nC, q2 = 24.9 nC

kq1/(r+0.174) = kq2/r

But this gives me one negative value for r, whereas I need two values.

 

[WatermelonPig]

Electric potential is a scalar.

 

[darkxponent]

since net potential is zero then

Potential due to q1 + potential due to q2=0

means equation is

kq1/(r+0.174) + kq2/r = 0 ; {not kq1/(r+0.174) = kq2/r}this will give two values of r one negative and one positive

 

[n_h0987]

I figured it out, thank you!

 

[rwisz]

My response:

The electric field magnitude at points where electric potential is zero depends on the distribution of charges in the system. In this case, we have two point charges, -14.3 nC and +24.9 nC, located 17.4 cm apart on the x-axis. The electric potential at any point on the x-axis can be calculated using the equation V = kq/r, where k is the Coulomb's constant, q is the charge, and r is the distance from the point charge. Since the electric potential at these two points is zero, we can set up the following equation: kq1/(r+0.174) = kq2/r. Solving for r gives us two values, one positive and one negative. This means that there are two points on the x-axis where the electric potential is zero. To find the electric field magnitude at these points, we can use the equation E = kq/r^2, where r is the distance from the point charge. Plugging in the values for q and r, we can calculate the electric field magnitude at both points. It is important to note that the electric field direction will be opposite at these two points, as the charges are of opposite signs.