Since the Van de Graaff generator is given to behave like a point charge, we can use the equation for Electric Field, given as
[tex]E = \frac{kQ}{r^{2} }[/tex]
where, [tex]k = \frac{1}{4\pi ε₀}[/tex] = [tex]8.99 X 10^{9} \frac{Nm^{2} }{C^{2} }[/tex]
r is the distance between the generator and the test charge = 0.75 m
E is the magnitude of the Electric Field Strength = [tex]4.5 X 10^{5} N/C[/tex]
Rearranging the equation, making Q the subject of the formula, we have
[tex]Q = \frac{E.r^{2} }{k}[/tex]
Plugging in the numerical values and simplifying them gets us
[tex]Q = 0.2815 X 10^{-4} \\[/tex]
Thus, the charge on the Van de Graaff generator is 28.16 μC.
