Question

A charged cloud system produces an electric field in the air near the earths surface. A particle of charge $ -2\times 10^{-9}C $ is acted on by a downward electrostatic force of $ 3\times 10^{-6}N$ when placed in this field. The gravitational and electrostatic force, respectively, exerted on a proton placed in this field are

• A. $1.64 \times 10^{-26} N , 2.4 \times 10^{-16} N$
• B. $1.64 \times 10^{-26} N , 1.5 \times 10^{-3} N$
• C. $1.56 \times 10^{-18} N , 2.4 \times 10^{-16} N$
• D. $1.5 \times 10^{-3} N , 2.4 \times 10^{-16} N$

Answer 1

Answer: (A)

For electrostatic force, $F=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{q_{1} q_{2}}{r^{2}} 3 \times 10^{-6}$
$=\frac{9 \times 10^{9} \times 2 \times 10^{-9} \times 2 \times 10^{-9}}{r^{2}}$
$\therefore r^{2}=\frac{9 \times 10^{9} \times 2 \times 10^{-9} \times 2 \times 10^{-9}}{3 \times 10^{-6}}$
Electrostatic force, when proton placed in this field,
$F=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{\text { Charge of partice } \times \text { charge of proton }}{r^{2}}$
$F=\frac{\frac{9 \times 10^{9} \times 2 \times 10^{-9} \times 1.6 \times 10^{-19}}{9 \times 10^{9} \times 2 \times 10^{-9} \times 2 \times 10^{-9}}}{3 \times 10^{-6}}$
$F=2.4 \times 10^{-16} N$
Similarly, we can calculate gravitational force
$=1.64 \times 10^{-26} N$