Pull It Away!

A point charge q q is placed at a distance l l from a infinite conducting sheet. The Work done in remove this charge from very far from sheet is given by

a b q 2 π ϵ 0 l \dfrac ab \dfrac {q^2}{π\epsilon_0 l}

Where a , b a, b are coprime positive integers.

Find a + b . a +b.


The answer is 17.

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1 solution

Work done d W = q E d l dW = qEdl

d W = q σ 2 ϵ 0 d l dW = q \dfrac {\sigma}{2\epsilon_0} dl

σ = Q 4 π r 2 {\color{#20A900}{\sigma = \dfrac {Q}{4πr^2}}}

W = 0 W d W = l q × q 2 × 4 π ϵ 0 ( 2 l ) 2 d l = q 2 2 × 4 π ϵ 0 × 2 1 l l = q 2 16 π ϵ 0 ( 1 l 1 ) = 1 16 q 2 π ϵ 0 l W = \displaystyle \int _0^W dW = \displaystyle \int _l^\infty q\times \dfrac {q}{2\times 4π\epsilon_0 (\sqrt{2}l)^2} dl = \dfrac {q^2}{2\times 4π\epsilon_0 \times 2} \begin{vmatrix} -\dfrac 1l \end{vmatrix}_l^\infty = \dfrac {q^2}{16π\epsilon_0}\begin{pmatrix} \dfrac 1l - \dfrac 1\infty \end{pmatrix} = \dfrac {1}{16} \dfrac {q^2}{π\epsilon_0l}

Therefore, a + b = 1 + 16 = 17 a + b = 1 + 16 = \boxed{17}

Is my solution even appropriate? Please let me know

Is it an infinite conducting sheet or an infinite charge sheet? The way the question is worded, it sounds like we're supposed to use the method of images. Also, if it is an infinite charge sheet, what is the area charge density σ \sigma ?

Steven Chase - 2 years, 4 months ago

Infinite conducting sheet... Can you post your solution?

A Former Brilliant Member - 2 years, 4 months ago

And about the σ \sigma Its the surface charge density. I mean thats how I was taught to solve. As its infinite conducting sheet, considering surface charge density and to solve the problem. Is it wrong?

A Former Brilliant Member - 2 years, 4 months ago

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