Answer :
Answer:
The excess charge is [tex]Q = 3.5 *10^{-7} \ C[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 45 \ cm = 0.45 \ m[/tex]
The potential of the surface is [tex]V = 14 \ kV = 14 *10^{3} \ V[/tex]
The radius of the sphere is
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.45}{2}[/tex]
[tex]r = 0.225 \ m[/tex]
The potential on the surface is mathematically represented as
[tex]V = \frac{k * Q }{r }[/tex]
Where k is coulomb's constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
given from the question that there is no other charge the Q is the excess charge
Thus
[tex]Q = \frac{V* r}{ k}[/tex]
substituting values
[tex]Q = \frac{14 *10^{3} 0.225}{ 9*10^9}[/tex]
[tex]Q = 3.5 *10^{-7} \ C[/tex]