Recall that for a finite rod of length L and total charge Q, the electric field for a point located a perpendicular distance r from the middle of the rod has a magnitude given by:
Consider the limit of large distances, where a finite charged rod is observed at a distance that is very large when compared to the length of the rod.
In this situation, the electric field best resembled that of a ...
a)point charge ... it goes like r-2
b) infinite line ... it goes like r-1
c) electric dipole ... it goes like r-3
d) infinite plane ... it is constant
none of the above ... it is zero
The finite rod of length L and charge Q looks more like a point source as the observer gets farther way.
The answer is a)
To determine the electric field at large distances from a finite charged rod, we need to consider the electric field contribution from each infinitesimally small element of the rod.
We start by calculating the electric field due to an infinitesimally small element located at a perpendicular distance r from the middle of the rod.
Considering a small element of length dl, the charge dq of this element is given by dq = Q/L * dl, where Q is the total charge of the rod and L is the length of the rod.
The electric field contribution dE from this small element at a point located at a distance r is given by Coulomb's Law as dE = k * dq / r^2, where k is the electrostatic constant.
Now, we integrate the contribution of all the infinitesimally small elements of the rod along its entire length to find the total electric field E at a distance r.
Given that the rod has a total charge Q, we integrate from -L/2 to L/2 (the full length of the rod). Integrating dq = Q/L * dl over this range gives dq = Q/L * dl = Q/L * dL, where dL is a small length element along the rod.
The distance from the middle of the rod to the element is given by r = sqrt(L^2/4 + L/2 - L/2) = sqrt(L^2/4), which can be approximated as r ≈ L/2 for large distances.
Substituting dq = Q/L * dL and r ≈ L/2 into the expression for dE, we get dE = k * Q/L * dL / (L/2)^2 = 4kQ/(πL^2) * dL.
Now, integrating dE over the entire length of the rod from -L/2 to L/2 gives:
E = ∫ dE = ∫ 4kQ/(πL^2) * dL from -L/2 to L/2.
Simplifying the integral, we find that E = 2kQ/(πL^2) * ∫ dL from -L/2 to L/2.
Integrating dL over this range gives E = 2kQ/(πL^2) * (L/2 - (-L/2)) = 2kQ/πL.
Thus, at large distances from the rod, the electric field best resembles that of an infinite line (option b) because it goes like r^-1.
Therefore, the correct answer is option b) infinite line.