A particle sits at a distance z away from the center of a bar of length L with a uniformly distributed charge Q. The electric field the particle experiences is given by E = (2kQ∕z)*(1∕√(4z^2 + L^2)) where k is a constant. (For those of you familiar with the laws of electricity, k = 1∕(4??0). We are going to simplify this electric field expression for two special cases. In the first case the particle is very close to the bar. This is the case where z ≪ L, so the unitless quantity z∕L is very small. (a) We can rewrite the electric field as E =(2kQ∕zL)(1∕√(4x^2 + 1)) ,where x = z∕L. Expand this function in a Maclaurin series up to the second power of x. Plug x = z∕L back in to get an approximate expression for E valid when the particle is near the bar. (b) The electric field at a distance z from an infinite bar of charge is E = 2k?∕z, where ? = Q∕L is the charge per unit length on the rod. The first term that you found should be equal to this expression, while the second term makes the approximation more accurate since the bar is not infinite. Explain why the second term has the sign that it does. In other words, would you expect the field from the finite bar to be larger or smaller than the field from an infinite bar with the same charge density, and does your result confirm your expectation? Our second special case is a particle very far from the bar. For this case L∕z will be very small. (c) In Part (a) we rewrote E by factoring L out of the square root to make it a function of z∕L. In a similar way, factor out 2z to make the square root a function of L∕z. (d) Create a Maclaurin series in L∕z up to the second power for the resulting function. (e) For this case your first term should reproduce the field from a point charge Q,namely E = kQ∕z2. The second term makes the expression more accurate since the bar is not an infinitely small point. Explain why the field from a bar of charge Q should be larger or smaller than the field from a point of charge Q and check that the sign of your second term matches this expectation.
I have parts a, b, and c answered.
Please Solve Parts d and e.
A particle sits at a distance z away from the center of a bar of length L with a uniformly distributed charge Q. The electric field the particle experiences is given by E = (2kQ∕z)*(1∕√(4z^2 + L^2)) where k is a constant. (For those of you familiar with the laws of electricity, k = 1∕(4??0).
We are going to simplify this electric field expression for two special cases. In the first case the particle is very close to the bar.
This is the case where z ≪ L, so the unitless quantity z∕L is very small.
(a) We can rewrite the electric field as E =(2kQ∕zL)(1∕√(4x^2 + 1)) ,where x = z∕L. Expand this function in a Maclaurin series
up to the second power of x. Plug x = z∕L back in to get an approximate expression for E valid when the particle is near the bar.
(b) The electric field at a distance z from an infinite bar of charge is E = 2k?∕z, where ? = Q∕L is the charge per unit length on the rod. The first term that you found should be equal to this expression, while
the second term makes the approximation more accurate since the bar is not infinite. Explain why the second term has the sign that it does. In other words, would you expect the field from the finite bar to
be larger or smaller than the field from an infinite bar with the same charge density, and does your result confirm your expectation?
Our second special case is a particle very far from the bar. For this case L∕z will be very small.
(c) In Part (a) we rewrote E by factoring L out of the square root to make it a function of z∕L. In a similar way, factor out 2z to make the square root a function of L∕z.
(d) Create a Maclaurin series in L∕z up to the
second power for the resulting function.
(e) For this case your first term should reproduce the field from a point charge Q,namely E = kQ∕z2. The second term makes
the expression more accurate since the bar
is not an infinitely small point. Explain why the field from a bar of charge Q should be larger or smaller than the field from a point of charge Q and check that the sign of your second term matches this
expectation.
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