Answer:
9.09 × 10⁶ volts
Explanation:
Given:
Diameter, d = 2.10 m
⇒ [tex]radius, r = \frac{d}{2}[/tex]
[tex]radius, r = \frac{2.10}{2}\\r = 1.05 \ m[/tex]
Charge, q = 1.06 mC = 1.06 × 10⁻³ C
Electric potential is the amount of work done in moving a unit charge from a reference point to a specific point inside electric field without producing an acceleration.
[tex]V = k\frac{q}{r} \ volts[/tex]
where
V is the electric potential in volt
k is the coulomb's constant 9 × 10⁹ Nm²/C²
q is the unit charge in Coulomb
r is radius of the sphere give in meter
∴ [tex]V = 9 * 10^{9} \ \frac{1.06 * 10^{-3}}{1.05} \ volts[/tex]
V = 9.085 × 10⁶ volts
V = 9.09 × 10⁶ volts
