Question Id: PH0000005
Question: What is the force between two small charged spheres having charges of \(2\times 10^{-7}C\) and \(3\times 10^{-7}C\) placed 30 cm apart in air?
Answer:
Force acting between \(q\) and \(Q\) is \(6\times 10^{-4}N\)
Explanation:
Given two charge \(q=2\times 10^{-7}C\) and \(Q=3\times 10^{-7}C\)
And they are separated by a distance \(r=30cm \Rightarrow r=0.3m\)
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| Figure-1 |
According to Coulomb's law
If two point charge \(q\) and \(Q\) are separated by a distance \(r\) then electrostatic force acting on them is \(F=\frac{1}{4\pi \epsilon }\frac{qQ}{r^{2}}\) ;[where \(\epsilon\) is electric permittivity and \(\epsilon =k\epsilon_{0}\) where k is dielectric constant and \(\epsilon_{0}\) is electric permittivity for air medium]
Now Force between \(q\) and \(Q\) is
\(F=\frac{1}{4\pi \epsilon _{0}}\frac{qQ}{r^{2}}\) ;[at air medium \(\epsilon =\epsilon_{0}\) as \(k=1\)]
\[\Rightarrow F=9\times 10^{9}\times \frac{2\times 10^{-7}\times 3\times 10^{-7}}{0.3^{2}}\]
\[\Rightarrow F=54\times \frac{10^{-6}}{0.09}\]
\[\Rightarrow F=6\times 10^{-4}\]
Hence, force acting between \(q\) and \(Q\) is \(6\times 10^{-4}N\). (Answer)
Theory related to this problem:
Condition for applying Coulomb's Law:
(i) Only applicable for point charges that means size of the point charge is negligible with respect to their distance.
(ii) Applicable for the distance larger than nuclear range ( \(\approx 10^{-15}m\) )
Coulomb's law (Theory): Attraction force or repulsion force between between two point charge is proportional to the product of the charge and inversely proportional to the square of the distance between the charge.
i.e \(F\alpha (q_{1}\times q_{2})\) and \(F\alpha \frac{1}{r^{2}}\)
So, \(F\alpha \frac{(q_{1}\times q_{2})}{r^{2}}\)
\(\Rightarrow F=k' \frac{(q_{1}\times q_{2})}{r^{2}}\) ; [where k' is proportionality constant]
In C.G.S Unit: \(k'=\frac{1}{k}\) [where k is electric permittivity]
Value of k=1 for air ( vaccum ) medium.
So, for C.G.S medium \(F=\frac{(q_{1}\times q_{2})}{r^{2}}\)
In S.I Unit: \(k'=\frac{1}{4\pi \epsilon }\) [ \(\epsilon\) permittivity of medium]
And \(\epsilon =k\epsilon _{0}\) where \(\epsilon _{0}\) is permittivity of air medium and k is dielectric constant.
So, \(F=\frac{1}{4\pi \epsilon _{0}}\times \frac{(q_{1}\times q_{2})}{r^{2}}\)
Value of \(\epsilon _{0}=8.854\times 10^{-12} C^{2}N^{-1}m^{-2}\)
And value of \(\frac{1}{4\pi \epsilon _{0}}=9\times 10^{9} Nm^{2}C^{-2}\)
Hence in S.I \(F=9\times 10^{9} \times \frac{(q_{1}\times q_{2})}{r^{2}}\) in N unit.
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