Two points are at a distance a and b apart (a<b) from left end of a uniformly charged rod as shown in the figure. The difference between the potentials at the two given points is proportional to
A
ba
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B
√ba
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C
ln(ba)
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D
ln{b(a−L)a(b−L)}
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Solution
The correct option is Dln{b(a−L)a(b−L)} Given that,
Length of the rod =L
Let the total charge on the rod is Q.
The electric potential (V) at a distance r from the end of a uniformly charged rod is given by
Vp=kQLln(r+Lr)
Here, we have
The electric potential at distance a from left end of rod is given by Va=kQLln(aa−L)
The electric potential at distance b from left end of rod is given by Vb=kQLln(bb−L)