Problem 1. Consider a conducting sphere of radius R and...

Problem 1. Consider a conducting sphere of radius $R$ and total charge $Q$ that is in the presence of the field generated by two point charges $q$ and $-q$ that are, respectively, at a distance $d$ and $-d$ from the center of the sphere with ($d > R$). ? Find the charge distribution on the conducting sphere. ? Consider now that the magnitude of the external charges is four times smaller than the charge on the conductor $4|q| = Q$. What form does the monopolar and electric dipoles contribution to the potential due to the whole arrangement take?

Transcript

00:01 Hello student, let's solve this question given that p bar is equal to kr square kr square r cap surface bound charge density sigma node is equal to p bar dot n cap for r is equal to r is equal to r now this is equal to kr square dot kr square kr square r kr square r kr square r kr square r kr kr square r kr kp dot r kp now this is equal to sigma not is equal to kr square sigma not is equal to kr square sigma not is equal to k r square volume volume volume volumes volume bounds charge volume bounds charge is equal to p b is equal to minus delta dot p now this is equal to p b is equal to p b is equal to 1 divided by r square r square, partial derivative with respect to d divided by r, r squared, dot pr.

01:06 Now this is equal to minus 1 divided by r squared, partial derivative d divided by dr is r square r square, r square, kr square, dot kr square, dot kr square.

01:21 Now solve this minus k divided by k divided by r square, partial derivative, d divided by d d divided by d r is equal to is the bracket is equal to r r to power 4 r to power 4 now solve this we will get minus k divided by r square multiply 4r cube multiply 4r cube now we will get p b is equal to minus 4 kr total bound total total bound charge is equal to is equal to sigma b b plus p b plus p b this is equal to kr square minus 4 kr this is equal to kr by 4 pi, 4 pi epsilon not are, now put the value 1 divided by 4 pi epsilon not r square r squared, some integration of 0 to r, 0 to r p b dv.

02:47 Now integrate this 1 divided by 4 pi epsilon node r square, integration of 0 to r, 0 to r, 4kr 4kr 4 pi r square d r 4 pi r square d r now further solve this this is equal to 1 divided by 4 pi epsilon node r square r square r square bracket minus 4 k 4 pi 4 pi integration of 0 to r r cube d r now e inside e inside e inside is equal to 1 divided by 4 pi xxyln node 4 pi epsilon not, bracket, minus 16 k pi, minus 16 k pi, divided by 4 pi epsilon 0 r squared r square, bracket, r to power 4 divided by 4, 0 to r...

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