Question

At what distance from a point charge of 8.0 μC would the electric potential be 4.2x104 V?

Answer 1

1.71 m

Step-by-step explanation

Given:

• The electric potential, E = 4.2*10⁴ V

,

• A point charge of q = 8*10⁻⁶ C

Unknown:

• The distance between the point charge and the electric potential, r.

The electric potential of a charge q at a distance r is given by,

`E=k\cdot(q)/(r)`

Where k is the Coulomb constant, with a value of approximately 8.99*10⁹ N*m²/C².

Solving for r,

`r=k\cdot(q)/(E)`

Replace with the known values and solve,

`r=8.99\cdot10^9\cdot(8\cdot10^(-6))/(4.2\cdot10^4)\approx1.71m`

Hence, the electric potential would be 1.71 m away from the point charge.