View the step-by-step solution to:

A positive charge per unit length lambda is distributed uniformly along a straight-line segment of length L.

2. A positive charge per unit length lambda is distributed
uniformly along a straight-line segment of length L.
(a) Determine the potential (chosen to be zero at
infinity) at a point P a distance y from one end of
the charged segment and in line with it. (b) Use the
result of (a) to compute the component of the electric
field at P in the y-direction (along the line). (c)
Find the electric field at P in the Y
direction by integrating along the line segment using the equation

dEy= (k dq/r^2) sin(theta)Jhat

sorry had to type the equation in words cause I don’t know how to do symbols….lol






1. Electric Quadrupole . The figure shows an electric quadrupole. It consists of two dipoles whose effects at external points do not quite cancel. Show that the value of the magnitude of the electric field E on the axis of the quadrupole for points a distance z from its center (assume z >> d) is given by


E = 3 Q / ( 4 pi epsilon0 z4 )
in which Q (=2 q d2) is known as the quadrupole moment of the charge distribution. [This is not the only way to get a quadrupole moment.]


This figure is actually supposed to be vertical with p being on top.






3 A battery has voltage E and internal resistance r. A variable loaded resistor R is connected across the terminals of the battery
b. Determine the value of R such that the potential difference across the terminals is a maximum
c. Determine the value of R that would maximize the current in the cicuit
d. Determine the value of R that would maximize the power delivered the load resistor. Choosing the load resistance for maximum power transfer is called impedance matching in general. This is important for getting the most boom out of the custom sound system in your car.

(Take derivative in order to maximize)



4 An initially uncharged capacitor C is fully charged by a constant voltage V in series with resistor R
a) Show that the final energy stored in the capacitor is half the energy supplied by the battery
b) By direct integration of I^2*R over the charging time, show that the internal energy dissipated by the resistor is also half the energy supplied by the battery


5. Amperes Law
A wire of radious R carries current I. The current density is given by J=J0 (1-r/R), where r is measured from the center of the wire and J0 is a constant. Use Amperes law to find:
A) The magnetic field inside the wire at a distance r from the central axis (r<R)
B) The magnetic field out side the wire a distance r from the central axis (r>R)

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