#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 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.

SHAIK AASIF AHAMED
6 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