Electric charge is distributed uniformly along a thin rod of length a , with tot

Question

Electric charge is distributed uniformly along a thin rod of length a, with total charge Q. Take the potential to be zero at infinity. Find the potential at the following points:

Part A

Find the potential at the point P, a distance x to the right of the rod.

Express your answer in terms of the given quantities and appropriate constants.

Part B

Find the potential at the point R, a distance y above the right-hand end of the rod.

Express your answer in terms of the given quantities and appropriate constants.

Part C

In part A, what does your result reduce to as x becomes much larger than a?

Express your answer in terms of the given quantities and appropriate constants.

Part D

In part B, what does your result reduce to as y becomes much larger than a?

Express your answer in terms of the given quantities and appropriate constants.

Explanation / Answer

Part A

Potential at a distance x from the right end of the rod be V(x)

Charge per unit length of the rod = Q/a

Let's take a tiny section of the rod of length dp at a position p from the right, then the tiny potential at x because of the charge on the tiny section will be

dV = k * (Q/a) * dp / (p + x) [using formula V = k * Q / R]

Now, to get the potential at x due to the whole rod we have to integrate the above equation.

V(x) = (kQ/a)*[Integral of dp/(p+x); between the limits of p = a to p = 0]

Which gives us,

V(x) = (kQ/a) * ln[(a+x)/x] = (kQ/a)*ln[1 +(a/x)]

V(x) = (kQ/a)*ln[1 +(a/x)]

Part B

Similarly, for point R, taking a small tiny rod and integrating it from a to 0 we get,

V(y) = (kQ/a)*[Integral of dp/[sqrt(p^2+y^2)] between the limits of p = a to p = 0]

So V(y) = (kQ/a)[ln[mod{p+sqrt(p^2+a^2)}] (for p = a) -n[mod{p+sqrt(p^2+a^2)} for p =0]

V(y) = (kQ/a)*ln[{a +sqrt(a^2+y^2)}/y] = (kQ/a)*ln[(a/y) + sqrt{1 + (a/y)^2]

V(y) = (kQ/a)*ln[(a/y) + sqrt{1 + (a/y)^2]

Part C

In part (a), if x becomes much larger than a

for x >> 0 tending to infinity V(x) = (kQ/a) * ln[1] = 0 [As x >> 0, a/x tends to 0]

If x becomes much larger than a, potential tends to 0 (ZERO)

Part D

In part (b), if y becomes much larger than a

for y >> 0 tending to infinity V(y) = (kQ/a) * ln[0 + sqrt{1 + 0}] = 0

If y becomes much larger than a, potential tends to 0 (ZERO)

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