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Question: An infinite number of charges, each of charge \(1\mu C\) are placed on the x-axis with coordinates \...

An infinite number of charges, each of charge 1μC1\mu C are placed on the x-axis with coordinates x=1,2,3..x=1,2,3\ldots \ldots ..\infty . If a charge of 1C1C is kept at the origin, then when what is the net force acting on 1C1C charge:

& A.9000N \\\ & B.12000N \\\ & C.24000N \\\ & D.36000N \\\ \end{aligned}$$
Explanation

Solution

Electrostatic force or coulomb’s inverse square law, which measures the forces between two charges, which at a distance rr from each other. Then the force is proportional to product of the charge and inversely proportional to the square of the distance between them. To remove the proportionality, a quantity kk is introduced.

Formula used:
F=14πϵ0Σqiqjrij2F=\dfrac{1}{4\pi\epsilon_{0}}\Sigma \dfrac{q_{i}q_{j}}{r_{ij}^{2}}

Complete answer:
The electrostatic force or the coulomb’s inverse square law was found by the French physicist, named Charles Augustin de Coulomb in the year 1785.
Let q1q_{1} and q2q_{2} be two point changes at a distant r12r_{12} between them, then we know that force due to a point charge is then given by: F12=14πϵ0q1q2r122F_{12}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{1}q_{2}}{r_{12}^{2}}.

If the force of interaction between the charges is attractive, then the value of FF is +ve and if the force of interaction between the charges is repulsive, then the value of FF is -ve .

Similarly, if nn number of charges is present, then the force due to them is given by:

F=14πϵ0Σqiqjrij2F=\dfrac{1}{4\pi\epsilon_{0}}\Sigma \dfrac{q_{i}q_{j}}{r_{ij}^{2}}, where qi,qjq_{i},q_{j} are the charges , rijr_{ij} is the distance between them respectively and ϵ0\epsilon_{0} is the permittivity of vacuum.
We can write 14πϵ0=k=9×109\dfrac{1}{4\pi\epsilon_{0}}=k=9\times10^{9} and is also known as coulomb’s constant.
Here, it is given that q1=1μCq_{1}=1\mu C and the rij=1,2,3..r_{ij}=1,2,3..\infty, then the FF on q1q_{1}, is given by:
F=k×106×[112+122+132+142+..12]F=k\times10^{-6} \times\left[\dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+..\dfrac{1}{\infty^{2}}\right]
Or, F=9×109×106×[11+14+19+116+..]F=9\times 10^{9}\times 10^{-6} \times\left[\dfrac{1}{1}+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+..\right]
Using the formula sum of infinite series, i.e. (a1r)\left(\dfrac{a}{1-r}\right), here a=1a=1 and r=14r=\dfrac{1}{4}
We get, F=9×103[1114]F=9\times 10^{3}\left[\dfrac{1}{1-\dfrac{1}{4}}\right]
Or, F=9×103[43]=12000NF=9\times 10^{3}\left[\dfrac{4}{3}\right]=12000N
Hence the answer is 12000N12000N

So, the correct answer is “Option B”.

Note:
The coulomb’s inverse square law is valid only for charges in rest. Also, the law can be extended to any number of charges. Note that FF is a vector quantity, i.e. it has both magnitude and direction. The resultant force is due to the superposition of forces due to charges, which is done by vector addition.