Q1. HCV example 18.... given: dipole moment = 3.4*10-30C-m... separation = 10-10m .. find mag. of charge..
simple q = D/d = 3.4*10-20C
doubt is how can q have charge less than proton ?
Q2. A positive charge q is placed in front of a conducting solid cube at a dist. d fro, its centre. Find the electric field at the centre of the cube due to the charges appearing on its surface...
11Read at your own risk [3]
for Q1 i think u have missed a 2
it is 2qd=D though that doesnt make much difference to your doubt[2]going nuts due to boards :(
sorry dint see the book
106im posting the full answer given in HCV..
If the charges on the two atoms are q and -q,
then q(1.0*10-10m) = 3.4*10-30C-m
or, q = 3.4*10-20 C
Note that this is less then the charge of a proton. Can you explain, how such a charge can appear on an atom
the one in bold is the exact answer given in HCV...
106i dont know... in HCV it is written that:
A combination of two charges +q and -q separated by a small distance d constitutes an electric dipole. the electric dipole moment of this combination is defined as a vector
p = qd
11very sorry i dunno what lead me to talk like that :(
sorry if it confused you too :(
106question edited it is HCl molecule ... toh H-bonding kahaan se aaya?? lagta hai Verma sir bhi bahut kuch chemistry padhte honge..[3]
11IN QUES 2
HINT
APPLY GAUSS LAW BY IMAGING NEW SURFACES AND BANK ON SYMMETRY
33For second it is q/4πε0d2 towards q
since cube is conducting so net field inside cube is zero...
so if field due to q is q/4πε0d2 away from q.. so net field due to cube should be q/4πε0d2
towards q to make it zero..
