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Question

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(aL)a(bL)}
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Solution

The correct option is D ln{b(aL)a(bL)}
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(aaL)
The electric potential at distance b from left end of rod is given by
Vb=kQLln(bbL)

Then the potential difference is,
ΔV=VbVa

ΔV=kQL{ln(bbL)ln(aaL)}
ΔV=kQL{ln(b(aL)a(bL))}
Vln(b(aL)a(bL))

Hence, option (d) is correct.

Why this question ?Tip: lnxlny=lnxy

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