Answer:
The potential is [tex]V = 153.659 \ V[/tex]
Explanation:
From the question we are told that
The diameter of the plastic sphere is [tex]d = 0.410 \ cm = 0.0041 \ m[/tex]
The magnitude of the charge is [tex]q = 35.0 pC = 35.0 *10^{-12} \ C[/tex]
The radius of the plastic sphere is mathematically evaluated as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{0.0041}{2}[/tex]
[tex]r = 0.00205 \ m[/tex]
The potential near the surface is mathematically represented as
[tex]V = \frac{k * q}{r }[/tex]
Where k is the Coulombs constant with value [tex]9 *10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
substituting values
[tex]V = \frac{9*10^9 * 35 *10^{-12}}{0.00205}[/tex]
[tex]V = 153.659 \ V[/tex]
