Two point charges of magnitude +40x10^-6C and -60x10^-6C are placed 50 cm apart. What is the distance from +40x10^-6C charge along the line joining the two point charges if the resultant potential is zero

potentials are scalar, so you can addthem

V=0=40E6/x-60E6/(.5-x)

solve for x

What is the exact formula to solve the question above?

To find the distance from the positive charge along the line joining the two point charges where the resultant potential is zero, we can use the concept of the electric potential due to point charges.

The electric potential due to a point charge is given by the formula:

V = k * q / r

Where:
- V is the electric potential
- k is the electrostatic constant (9 x 10^9 Nm^2/C^2)
- q is the magnitude of the charge
- r is the distance from the charge

In this case, there are two point charges, +40x10^-6C and -60x10^-6C.

Let's assume the positive charge (+40x10^-6C) is placed at a distance "x" from it along the line joining the two point charges.

The electric potential due to the positive charge will be:

V1 = k * q1 / r1

And the electric potential due to the negative charge will be:

V2 = k * q2 / r2

For the resultant potential to be zero, V1 + V2 = 0.

Substituting the values, we get:

(k * q1 / r1) + (k * q2 / r2) = 0

(40x10^-6C) / r1 - (60x10^-6C) / r2 = 0 [Substituting values for q1 and q2]

Now, we know that the distance between the two charges is 50 cm or 0.5 m. So, we have:

r1 + r2 = 0.5m

We need to find the value of r1, so it can be expressed in terms of r2.

Let's rearrange the equation to solve for r1:

r1 = (0.5m * r2) / (r2 - 0.5m)

Now, we can substitute this value of r1 in the equation (40x10^-6C) / r1 - (60x10^-6C) / r2 = 0 and solve for r2.

Once we have the value of r2, we can substitute it back into the equation r1 + r2 = 0.5m to find the value of r1.