ANSWER
1.71 m
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,
[tex]E=k\cdot\frac{q}{r}[/tex]Where k is the Coulomb constant, with a value of approximately 8.99*10⁹ N*m²/C².
Solving for r,
[tex]r=k\cdot\frac{q}{E}[/tex]Replace with the known values and solve,
[tex]r=8.99\cdot10^9\cdot\frac{8\cdot10^{-6}}{4.2\cdot10^4}\approx1.71m[/tex]Hence, the electric potential would be 1.71 m away from the point charge.
