Answer:
+1.33 × [tex]10^{-7}[/tex] C and -1.33 × [tex]10^{-7}[/tex] C respectively.
Explanation:
Electric potential (V) is the work done in moving a unit positive charge from infinity to a reference point within an electric field. It is measured in volts.
V = [tex]\frac{kq}{r}[/tex] ............. 1
where: k is a constant = 9 × [tex]10^{9}[/tex] N[tex]m^{2} C^{-2}[/tex], q is the charge and r is the distance between the charges.
From equation 1,
q = [tex]\frac{Vr}{k}[/tex] ............... 2
The charge on each ball can be determined as;
given that; V = 1.2 × [tex]10^{3}[/tex], k = 9 × [tex]10^{9}[/tex] N[tex]m^{2} C^{-2}[/tex] and r = 1.00 m.
From equation 2,
q = [tex]\frac{1.2*10^{3} * 1.0}{9*10^{9} }[/tex]
= 1.33 × [tex]10^{-7}[/tex] C
Thus, the charge on the first ball is +1.33 × [tex]10^{-7}[/tex] C, while the charge on the second ball is -1.33 × [tex]10^{-7}[/tex] C.
