The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Question Id: PH0000006 

Question: The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Answer:

(a) Distance between the two spheres is 0.12m.

(b) Force on the second sphere due to the first sphere is save and i.e  0.2 N.

Explanation:

Given, 

two small charge \(q_{1}=0.4\mu C=0.4\times 10^{-6}C=4\times 10^{-7}C\) ,  \(q_{2}=-0.8\mu C=-0.8\times 10^{-6}C=-8\times 10^{-7}C\)  

and force(F)  between them is 0.2 N.

Distance between the sphere of charges is r (say)

Figure-1: force between 0.4 µC and –0.8 µC

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 \(k=1\)]

\[\Rightarrow 0.2=9\times 10^{9}\times \frac{(4\times 10^{-7}\times 8\times 10^{-7})}{r^{2}}\]

\[\Rightarrow r^{2}=9\times 10^{9}\times \frac{(4\times 10^{-7}\times 8\times 10^{-7})}{0.2}\]

\[\Rightarrow r^{2}=\frac{288\times 10^{9-14}}{0.2}\]

\[\Rightarrow r^{2}=\frac{288\times 10^{-5}}{2\times 10^{-1}}\]

\[\Rightarrow r^{2}=\frac{288}{2} \times 10^{-4}\]

\[\Rightarrow r^{2}=144 \times 10^{-2}\]

\[\Rightarrow r=\pm 12 \times 10^{-2}\]

\[\Rightarrow r=\pm 0.12\]

So, \(r=0.12m\) as distance is always positive. 

According to Coulomb's Law, electrostatic force of attraction or repulsion is same but opposite for each point charge. This is the Newtonian property of Coulomb's force. Hence, one can say that force on the second sphere due to the first sphere is save and i.e  0.2 N.

(a) Distance between the two spheres is 12m.

(b) Force on the second sphere due to the first sphere is save and i.e  0.2 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.

From vector form Coulomb's Law it can say that it is Newtonian in nature i.e \(\overrightarrow F_{12}=-\overrightarrow F_{21}\).