b
(b) For equilibrium \(mg = qE\)
\(1.96 \times {10^{ - 15}} \times 9.8 = q \times \,\left( {\frac{{800}}{{0.02}}} \right)\)
\(==>\) \(q = \frac{{1.96 \times {{10}^{ - 15}} \times 9.8 \times 0.02}}{{800}}\)
\(==>\) \(n \times 1.6 \times {10^{ - 19}} = \frac{{1.96 \times {{10}^{ - 15}} \times 9.8 \times 0.02}}{{800}}\)
\(n = 3\).