1. Key K is connected in turn to each of the contacts over short identical time intervals so that change in charge on capacitor is small. Final charge on cap is:
1)2.5 C
2)5 C
3)0.5 C
2. There is a dielectric rod b/w charges q1 and q2 and now the rod is removed. Net force on q1 now:
Increases
decreases
Remains same
Both 1 and 2 depending on nature of charges
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4 Answers
62The first question is a very good question..
Think of a charge q there.. on the capacitor...
In the time dt, the charge on the capacitor will be given by
3=dq/dt+q/c
3dt = dq + qdt/c
In the next time dt, the charge dq should flow out.. so here, the dq should be -ve of the dq in the first loop.
2=dq/dt+q/c
2dt=dq+qdt/c
so,
3dt+2dt=2qdt/c
so q=5/2C
62If we ignore the end effect, the electric field outside the dielectric does not change due to the introduction of a dielectric..
So outside the dielectric, there is no change in the electric field..
So there will be no change in the force on the two rods.
1thanks for the answers bhaiya..
both the qs are from the 'success magnet' question bank of Aaakash Institute.
First one is correct
But ans to q2 is given as d)