00:01
Hi, here in this given problem there are three different questions.
00:05
In the first one that is question number 10, the two conducting spheres in contact with each other, the smaller one and the bigger one and a charged rod, positively charged rod brought close to the sphere a.
00:30
So, due to induction the sphere a will become negatively charged and sphere b will become positively charged.
00:39
Now, when b is shifted a little towards left, the positive charge will remain over sphere b and negative charge will remain over sphere a.
00:52
So, we can say finally, sphere a will be negatively charged and sphere b it will be positively charged.
01:06
Hence, option c is correct in this problem.
01:14
Then the next one, question number 11, out of three identical spheres, charge over sphere a, qa, minus 2 .0 microcoulomb over b, qb is equal to minus 6 .0 microcoulomb and over c this is plus 4 .0 microcoulomb.
01:37
First of all, a and b are touched and separated, then the redistributed charge over them will become qa dash is equal to qb dash and that will be given by qa plus qb divided by 2, means minus 2 .0 microcoulomb minus 6 .0 microcoulomb divided by 2 and that is minus 4 .0 microcoulomb over each one of them...