(a) What is the electric field \(5.00{\rm{ m}}\) from the center of the terminal of a Van de Graaff with a \(3.00{\rm{ mC}}\) charge, noting that the field is equivalent to that of a point charge at the center of the terminal? (b) At this distance, what force does the field exert on a \({\rm{2}}{\rm{.00 \mu C}}\) charge on the Van de Graaff’s belt?

Van de Graaff generator is machine which produces large electrostatic potential difference at low current levels.It accumulates the electric charge on a hollow metal globe mounted at the top of an insulated column by using moving belt.

The electric field is defined as the force acting per unit positive charge when placed in the space around another charge. The expression for the electric field is given as,

\(E = \frac{{Kq}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant \(\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the magnitude of the charge \(\left( {q = 3.00{\rm{ }}mC} \right)\), and \(r\) is the distance of the point of consideration \(\left( {r = 5.00{\rm{ }}m} \right)\).

Substituting all known values,

\(\begin{array}{c}E = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {3.00{\rm{ }}mC} \right)}}{{{{\left( {5.00{\rm{ }}m} \right)}^2}}}\\ = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {3.00{\rm{ }}mC} \right) \times \left( {\frac{{{{10}^{ - 3}}{\rm{ }}C}}{{1{\rm{ }}mC}}} \right)}}{{{{\left( {5.00{\rm{ }}m} \right)}^2}}}\\ = 1.08 \times {10^6}{\rm{ N/C}}\end{array}\)

Hence, the electric field \(5.00{\rm{ }}m\) from the center of the terminal of a Van de Graaff is \(1.08 \times {10^6}{\rm{ N/C}}\).

The electric field is defined as,

\(E = \frac{F}{{{q_o}}}\)

Here, \(E\) is the electric field \(\left( {E = 1.08 \times {{10}^6}{\rm{ }}N/C} \right)\), \(F\) is the force on the test charge, and \({q_o}\) is the test charge \(\left( {{q_o} = 2.00{\rm{ }}\mu C} \right)\).

The expression for the force is,

\(F = {q_o}E\)

Substituting all known values,

\(\begin{array}{c}F = \left( {2.00{\rm{ \mu C}}} \right) \times \left( {1.08 \times {{10}^6}{\rm{ N/C}}} \right)\\ = \left( {2.00{\rm{ \mu C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ }}C}}{{1{\rm{ \mu C}}}}} \right) \times \left( {1.08 \times {{10}^6}{\rm{ N/C}}} \right)\\ = 2.16{\rm{ N}}\end{array}\)

Hence, the force on the charge on the Van de Graaff’s belt is \(2.16{\rm{ N}}\).