Question

A charged cloud system produces an electric field in the air near the earth’s surface. A particle of charge $ - 2 \times {10^{ - 9}}C$ is acted on by a downward electrostatic force of $3 \times {10^{ - 16}}N$ 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

Hint: When a charge particle enter into a uniform electric field then electrostatic force on the charge particle is given as ${F_e} = qE$ and the gravitational force between 2 massive bodies having masses ${m_1}$ & ${m_2}$ is given as ${F_g} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Where
r $ = $ distance between ${m_1}$ & ${m_2}$

Complete step by step answer:
When a charge particle enter into a uniform electric field then electrostatic force on the charge particle is given as ${F_e} = qE$ and the gravitational force between 2 massive bodies having masses ${m_1}$ & ${m_2}$ is given as ${F_g} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Where
r $ = $ distance between ${m_1}$ & ${m_2}$
Step by step solution :
Given that a particle of charge $ - 2 \times {10^{ - 9}}C$ is acted on by a charged cloud by a downward electrostatic force of $3 \times {10^{ - 6}}N$. So, the electric field of charged cloud is given as –
$F = qE$
$E = \dfrac{F}{q} = \dfrac{{3 \times {{10}^{ - 6}}}}{{2 \times {{10}^{ - 9}}}}$
$E = \dfrac{3}{2} \times {10^3}$
$E = 1.5 \times {10^3}N/C$
So due to this electric field, the electric force exerted on a proton is
${F_e} = {q_p}E$
Here ${q_p} = $ charge of proton $(1.67 \times {10^{ - 19}}C)$
So, force on proton is
${F_e} = 1.6 \times {10^{ - 19}} \times 1.5 \times {10^3}$
${F_e} = 2.4 \times {10^{ - 16}}N$ …..(1)
The expression for gravitational force between earth and a proton (at earth surface) is given as
${F_g} = \dfrac{{G{m_p}{m_e}}}{{{r^2}}}$
Here
${m_p} = 1.67 \times {10^{ - 27}}kg$
${m_e} = 5.97 \times {10^{24}}kg$
r $ = $ distance between proton & earth centre point i.e., radius of earth $({R_e})$
$r = {R_c} = 6400km = 6400 \times {10^3}m$
and G $ = $ gravitational constant
$ = 6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$
So, ${F_g} = \dfrac{{6.67 \times {{10}^{ - 11}} \times 1.67 \times {{10}^{ - 27}} \times 5.97 \times {{10}^{24}}}}{{6400 \times 6400 \times {{10}^3} \times {{10}^3}}}$
$\Rightarrow {F_g} = \dfrac{{6.67 \times 1.67 \times 5.97 \times {{10}^{ - 11}} \times {{10}^{ - 27}} \times {{10}^{24}}}}{{64 \times 64 \times {{10}^3} \times {{10}^3} \times {{10}^2} \times {{10}^2}}}$
$\Rightarrow {F_g} = \dfrac{{66.4992}}{{4096}} \times \dfrac{{{{10}^{ - 14}}}}{{{{10}^{10}}}}$
$\Rightarrow {F_g} = 0.0162351 \times {10^{ - 14}} \times {10^{ - 10}}$
$\Rightarrow {F_g} = 0.016235 \times {10^{ - 24}}$
$\Rightarrow {F_g} \simeq 1.64 \times {10^{ - 26}}N$ …..(2)

So, the correct answer is “Option A”.

Note:
Many times, they can ask the ratio of gravitational force and electrostatic force.
So, ${F_g} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
${F_e} = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}$
$\dfrac{{{F_g}}}{{{F_e}}} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}} \times \dfrac{{{r^2}}}{{k{q_1}{q_2}}}$
$\dfrac{{{F_g}}}{{{F_e}}} = \dfrac{{G{m_1}{m_2}}}{{k{q_1}{q_2}}}$