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Q.NO-1)There are two large parallel sheets S1 and S2 carrying surface charge densities s1 and s2 respectively (s1>s2) placed at a distance d apart in vacuum. Find the work done by the electric field in moving a point charge q at a distance a ( a Q.NO2)There are two large parallel sheets S1 and S2 carrying surface charge densities s1 and s2 respectively (s1>s2) placed at a distance d apart in vacuum. Find the work done by the electric field in moving a point charge q at a distance a ( a A rock is 1.5 x 109 years old. The rock contains 238U, which disintegrates to form 206Pb. Assume that there was no 206Pb in the rock initially and it is the only stable product formed by the decay , calculate the ratio of number of nuclei of 206Pb to that of 238U in the rock. Half life of 238U is 4.5 x 109 years.

Q.NO-1)There are two large parallel sheets S1 and S2 carrying surface charge densities s1 and s2 respectively (s1>s2) placed at a distance d apart in vacuum. Find the work done by the electric field in moving a point charge q at a distance a ( a < d) from S1 towards S2 along a line making an angle p/4 with the normal to the plates.




Q.NO2)There are two large parallel sheets S1 and S2 carrying surface charge densities s1 and s2 respectively (s1>s2) placed at a distance d apart in vacuum. Find the work done by the electric field in moving a point charge q at a distance a ( a < d) from S1 towards S2 along a line making an angle p/4 with the normal to the plates.


A rock is 1.5 x 109 years old. The rock contains 238U, which disintegrates to form 206Pb. Assume that there was no 206Pb in the rock initially and it is the only stable product formed by the decay , calculate the ratio of number of nuclei of 206Pb to that of 238U in the rock. Half life of 238U is 4.5 x 109 years.

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1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
9 years ago
Hello student,
1)Electric field in the space between the plates towards right
E=\small (\sigma _{1}-\sigma _{2})/2\varepsilon _{0}
So work done W=Fs cos\small \Theta
=qEacos (\small \pi /4)
=\small qa(\sigma _{1}-\sigma _{2})/2\sqrt{2}\varepsilon _{0}
2) Initial number of nuclei N0 0
Final number of nuclei N N0-N
N=N0(1/2)nwhere n is number of half lives=(1/3) in this case
(N/N0)=1/21/3so (N0-N)/N=0.259
Thanks and Regards
Shaik Aasif
askIITians faculty

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