Electric Field Intensity at the Origin!

An infinite number of charges each of magnitude q q are placed on the x x - axis \text {axis} at the distance of 1 , 2 , 4 , 8 , 16 , 1,2,4,8,16,\ldots m \text{m} respectively. Find the intensity of electric field at the origin.

If the intensity can be expressed in the form q n π ε o \dfrac { q }{ n\pi { \varepsilon }_{ o } } , find Γ ( n + 1 ) \Gamma (n+1) .

  • Assume the medium to be vacuum.

  • Here, Γ ( n ) \Gamma (n) and ε o { \varepsilon }_{ o } represent the Gamma function and permittivity of free space respectively.


The answer is 6.

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1 solution

Rishabh Jain
Apr 18, 2016

Using formula of field for a point charge E = q 4 π ϵ 0 r 2 E=\dfrac{q}{4\pi\epsilon_0r^2} and Superposition principle , the net field by all the charges is :

E n e t = q 4 π ϵ 0 ( 1 + 1 2 2 + 1 4 2 + 1 8 2 ) E_{net}=\dfrac{q}{4\pi\epsilon_0}\left(1+\dfrac1{2^2}+\dfrac{1}{4^2}+\dfrac{1}{8^2}\cdots\right)

= q 4 π ϵ 0 ( 1 + 1 2 2 + 1 2 4 + 1 2 6 ) (Infinite GP) =\dfrac{q}{4\pi\epsilon_0}\underbrace{\left(1+\dfrac1{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}\cdots\right)}_{\color{#D61F06}{\text{(Infinite GP)}}}

= q 4 π ϵ 0 ( 1 1 1 / 4 ) \large =\dfrac{q}{4\pi\epsilon_0}\left(\dfrac{1}{1-1/4}\right)

= q 3 π ϵ 0 \large =\dfrac{q}{\color{#20A900}{3}\pi\epsilon_0}

Γ ( 3 + 1 ) = 6 \therefore\huge\Gamma (\color{#20A900}{3}+1)=\boxed 6

I have a doubt problem on Coulomb's Law. If I ask it here, could you post the solution as a comment?

Swapnil Das - 5 years, 1 month ago

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You can post it as a note maybe! So that more people can come to know abt it and maybe post their different perspectives...

Rishabh Jain - 5 years, 1 month ago

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Oh, good idea. Please comment on the note if possible.

Swapnil Das - 5 years, 1 month ago

Yeah Same way!!!

Kaustubh Miglani - 5 years, 1 month ago

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Great... :-)

Rishabh Jain - 5 years, 1 month ago

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