the distance between two negatively charged dust particle is 1mm and they repel each other with a force of 4×10^'-5 N. if the change don one particle is 4 times that of the other then find out the number of electrons charging them separately.
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Answered by
147
According to Coulomb forces:
F= K q1 q2/r²-----------------(1)
where K= 1/4π ε₀ = 9x10⁹ Nm²/C²
F=4x10⁻⁵ N
according to given : q1=4q2
r=1mm=10⁻³m
Substituting all these values in equation 1 we get,
4x10⁻⁵ =9x10⁹ xq1 x4 q1/(10⁻³)²
4x10⁻⁵ =9x10⁹ xq1 x4 q1/10⁻⁶
10⁻⁵=9x10¹⁵ xq1²
q1² =10⁻⁵/9x 10 ¹⁵
q1² =10⁻²⁰/9
q1=(1/3 ) x 10⁻¹⁰ C
As we know the equation Q=ne
where n is number of electrons and e is charge on electrons.
But in the question the value of e is not given. If i take e =5/3 x 10⁻¹⁹ C
then n=q1/e=(1/3)(10⁻¹⁰x 10¹⁹)/5x 10⁻¹⁹
=10⁹/5
number of electrons on charge q1= 2x10⁸
Number of electrons on charge q2=4q1=4x10⁸
F= K q1 q2/r²-----------------(1)
where K= 1/4π ε₀ = 9x10⁹ Nm²/C²
F=4x10⁻⁵ N
according to given : q1=4q2
r=1mm=10⁻³m
Substituting all these values in equation 1 we get,
4x10⁻⁵ =9x10⁹ xq1 x4 q1/(10⁻³)²
4x10⁻⁵ =9x10⁹ xq1 x4 q1/10⁻⁶
10⁻⁵=9x10¹⁵ xq1²
q1² =10⁻⁵/9x 10 ¹⁵
q1² =10⁻²⁰/9
q1=(1/3 ) x 10⁻¹⁰ C
As we know the equation Q=ne
where n is number of electrons and e is charge on electrons.
But in the question the value of e is not given. If i take e =5/3 x 10⁻¹⁹ C
then n=q1/e=(1/3)(10⁻¹⁰x 10¹⁹)/5x 10⁻¹⁹
=10⁹/5
number of electrons on charge q1= 2x10⁸
Number of electrons on charge q2=4q1=4x10⁸
Answered by
20
Answer:
Explanation:then n=q1/e=(1/3)(10⁻¹⁰x 10¹⁹)/5x 10⁻¹⁹
=10⁹/5
Isme (1/3) kha gya
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