Question

A charge q1 = 2.42 µC is at a distance d = 1.63 m from a second charge q2 = −5.37 µC.

Find the electric potential at a point A between the two charges that is 3d/4 from q1.

Answer 1

The electric potential at point A, located between the two charges, is -1.118 volts.

Step-by-step explanation:

To find the electric potential at point A between the two charges, we can use the formula for electric potential:

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

Here, k is the electrostatic constant (k = 9 × 10^9 Nm²/C²), q1 is the charge of the first charge (2.42 µC), r1 is the distance between point A and the first charge (3d/4 = 1.22 m), q2 is the charge of the second charge (-5.37 µC), and r2 is the distance between point A and the second charge (d/4 = 0.4075 m).

Substituting the values into the formula:

V = (9 × 10^9 Nm²/C²) * (2.42 µC / 1.22 m - 5.37 µC / 0.4075 m)

Simplifying the equation, we get:

V = (9 × 10^9 Nm²/C²) * (1.9836 × 10^-6 C/m - 1.316 × 10^-5 C/m)

V = -1.118 V

Therefore, the electric potential at point A is -1.118 volts.