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Question: A charged cloud system produces an electric field in the air near the earth’s surface. A particle of...

A charged cloud system produces an electric field in the air near the earth’s surface. A particle of charge −2×10−9C - 2 \times {10^{ - 9}}C is acted on by a downward electrostatic force of 3×10−16N3 \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) 1.64×10−26N,2.4×10−16N1.64 \times {10^{ - 26}}N,2.4 \times {10^{ - 16}}N
(B) 1.64×10−26N,1.5×103N1.64 \times {10^{ - 26}}N,1.5 \times {10^3}N
(C) 1.56×10−18N,2.4×10−16N1.56 \times {10^{ - 18}}N,2.4 \times {10^{ - 16}}N
(D) 1.5×103N,2.4×10−16N1.5 \times {10^3}N,2.4 \times {10^{ - 16}}N

Explanation

Solution

When a charge particle enter into a uniform electric field then electrostatic force on the charge particle is given as Fe=qE{F_e} = qE and the gravitational force between 2 massive bodies having masses m1{m_1} & m2{m_2} is given as Fg=Gm1m2r2{F_g} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}
Where
r == distance between m1{m_1} & m2{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 Fe=qE{F_e} = qE and the gravitational force between 2 massive bodies having masses m1{m_1} & m2{m_2} is given as Fg=Gm1m2r2{F_g} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}
Where
r == distance between m1{m_1} & m2{m_2}
Step by step solution :
Given that a particle of charge −2×10−9C - 2 \times {10^{ - 9}}C is acted on by a charged cloud by a downward electrostatic force of 3×10−6N3 \times {10^{ - 6}}N. So, the electric field of charged cloud is given as –
F=qEF = qE
E=Fq=3×10−62×10−9E = \dfrac{F}{q} = \dfrac{{3 \times {{10}^{ - 6}}}}{{2 \times {{10}^{ - 9}}}}
E=32×103E = \dfrac{3}{2} \times {10^3}
E=1.5×103N/CE = 1.5 \times {10^3}N/C
So due to this electric field, the electric force exerted on a proton is
Fe=qpE{F_e} = {q_p}E
Here qp={q_p} = charge of proton (1.67×10−19C)(1.67 \times {10^{ - 19}}C)
So, force on proton is
Fe=1.6×10−19×1.5×103{F_e} = 1.6 \times {10^{ - 19}} \times 1.5 \times {10^3}
Fe=2.4×10−16N{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
Fg=Gmpmer2{F_g} = \dfrac{{G{m_p}{m_e}}}{{{r^2}}}
Here
mp=1.67×10−27kg{m_p} = 1.67 \times {10^{ - 27}}kg
me=5.97×1024kg{m_e} = 5.97 \times {10^{24}}kg
r == distance between proton & earth centre point i.e., radius of earth (Re)({R_e})
r=Rc=6400km=6400×103mr = {R_c} = 6400km = 6400 \times {10^3}m
and G == gravitational constant
=6.67×10−11m3kg−1s−2= 6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}
So, Fg=6.67×10−11×1.67×10−27×5.97×10246400×6400×103×103{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}}}
⇒Fg=6.67×1.67×5.97×10−11×10−27×102464×64×103×103×102×102\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}}}
⇒Fg=66.49924096×10−141010\Rightarrow {F_g} = \dfrac{{66.4992}}{{4096}} \times \dfrac{{{{10}^{ - 14}}}}{{{{10}^{10}}}}
⇒Fg=0.0162351×10−14×10−10\Rightarrow {F_g} = 0.0162351 \times {10^{ - 14}} \times {10^{ - 10}}
⇒Fg=0.016235×10−24\Rightarrow {F_g} = 0.016235 \times {10^{ - 24}}
⇒Fg≃1.64×10−26N\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, Fg=Gm1m2r2{F_g} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}
Fe=kq1q2r2{F_e} = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}
FgFe=Gm1m2r2×r2kq1q2\dfrac{{{F_g}}}{{{F_e}}} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}} \times \dfrac{{{r^2}}}{{k{q_1}{q_2}}}
FgFe=Gm1m2kq1q2\dfrac{{{F_g}}}{{{F_e}}} = \dfrac{{G{m_1}{m_2}}}{{k{q_1}{q_2}}}