Thank you Damon for answering my first questions fast and accurately. Do you or anybody else know how to do these harder Gauss' Law Problems?
1. A solid metal sphere of radius R has charge +2Q. A hollow spherical shell of radius 3R placed concentric with the first sphere has net charge -Q.
a) Describe the electric field lines both inside and outside the spheres (as if you had to draw field lines on the chart-no calculations).
b) Use Gauss' law to find an expression for the magnitude of the electric field between the spheres at a distance r from the center of the inner sphere (R<r<3R)
c) Calculate the potential difference between the two spheres.
d) The spheres are then connected by a conducting wire for long enough that the charges can redistribute. What is the final distribution charge on the two spheres?
e) The wire connecting the spheres is removed and the outside sphere is grounded. What is the charge distribution on the two spheres?
2. A long, insulating cylinder has a uniform charge density, ro=-2*mu*C/m^3, and a radius, R=15cm.
a) What is the total charge on a 40cm length of that cylinder?
b) Use Gauss' Law to find the electric field both inside and outside of the cylinder as a function of the distance from the center of the cylinder, r.
c) Set up an integral to evaluate the potential at the surface of the cylinder (dont solve).
An isolated (ungrounded) conducting shell of inside radius 30cm and outside radius 40cm is now snapped together around the insulating cylinder.
d) What is the effect (qualitative) on the electric field at r=50cm?
e) What is the effect (qualitative) on the potential at the surface of the insulator.
a)
inside the inner sphere, a closed spherical surface encloses no charge, so E is 0
between spheres, the outer sphere is irrelevant, only the 2 Q matters
E drops off with 1/r^2, the surface area of spherical surface.
outside sphere 2
we surround a total of 2Q-Q = 1 Q
field again drops off as 1/r^2
b)
R<r<3R
E = 2 Q/(4 pi r^2 eo)
c)
integral E dr = [Q/(2 pi eo)] dr/r^2
= [Q/(2 pi eo)] [-1/r] at 3R - at 1R
= [Q/(2 pi eo)] [ -1/3R +1/R ]
= (2/3)[Q/(2 pi eo R)]
= Q/(3 pi eo R)
d) the E between spheres must be zero or there would be a potential difference between spheres, so all charges must move to the outer sphere, net 1 Q
e) all gone.
Damon, you are amazing...
Tell me if you get the second one please, I'm still having trouble.
In the meantime I temporarily took over your role answering all the math questions =].
2. A long, insulating cylinder has a uniform charge density, ro=-2*mu*C/m^3, and a radius, R=15cm.
a) What is the total charge on a 40cm length of that cylinder?
volume of .4 m length = pi r^2 L = pi (.15)^2 (.4) = .02827 m^3
Q = ro * volume = -.05655 *10^-6
= -5.66*10^-8 C
b) Use Gauss' Law to find the electric field both inside and outside of the cylinder as a function of the distance from the center of the cylinder, r.
E
inside
Q enclosed = ro pi r^2 L
area of Gauss surface = 2 pi r L
E = Qenclosed/area eo = ro pi r^2 L/2 pi r L eo
= ro r/2 eo
outside
E = Q at R length L / 2 pi r eo L
= ro pi R^2 L / 2 pi r eo L
= ro (R^2/r)/2 eo
E *area = Qenclosed / 4 pi r^2 eo
c) Set up an integral to evaluate the potential at the surface of the cylinder (dont solve).
well, I suppose we call it 0 at the center so
integral E dr = integral inside
= (ro /2 eo) r dr
An isolated (ungrounded) conducting shell of inside radius 30cm and outside radius 40cm is now snapped together around the insulating cylinder.
d) What is the effect (qualitative) on the electric field at r=50cm?
still surround the same old charge, -Q moves inside, + Q moves to outer surface. No effect
e) What is the effect (qualitative) on the potential at the surface of the insulator. No effect
Now if you ground that conducting shell, that kind of defines your zero of potential there, whereas I defined it at the center of the problem. That would change Voltages but not voltage differences.
Thank you so much Damon!
To solve these harder Gauss' Law problems, you will need to understand the basic concepts of Gauss' Law and be comfortable with using mathematical equations and integrals. Here's how you can approach each problem:
1. A solid metal sphere of radius R has charge +2Q. A hollow spherical shell of radius 3R placed concentric with the first sphere has net charge -Q.
a) To describe the electric field lines both inside and outside the spheres, you can use Gauss' Law to visualize the field lines. Inside the solid sphere, the electric field lines would be radially inward, pointing towards the center. Inside the hollow shell, there would be no electric field lines since the net charge inside the shell is zero. Outside the hollow shell, the field lines would be radially outward, away from the center.
b) To find an expression for the magnitude of the electric field between the spheres at a distance r from the center of the inner sphere (R<r<3R), you can apply Gauss' Law. Choose a Gaussian surface in the form of a concentric spherical shell between the two spheres and calculate the total electric flux through that surface. Use the fact that the net charge enclosed by the Gaussian surface is Q (from the inner solid sphere) since the charge of the hollow shell is canceled out. Divide the total electric flux by the surface area of the Gaussian shell to find the expression for the magnitude of the electric field.
c) To calculate the potential difference between the two spheres, you can integrate the electric field along a path from the surface of the inner sphere to the surface of the outer shell. Use the expression for the magnitude of the electric field you found in part b) and integrate it with respect to distance.
d) After connecting the spheres with a conducting wire, the charges will redistribute until they reach a state of equilibrium. Since the spheres are connected, they will share charge until they have the same potential. Use the fact that the potential is the same throughout the connected spheres and set up a relation between the initial charges (2Q and -Q) and the final charges on the spheres.
e) When the wire connecting the spheres is removed and the outside sphere is grounded, the potential of the grounded sphere is fixed at zero. From the previous part, you can determine the final charge distribution on the two spheres by noting that the charge on the grounded sphere redistributes to make its potential zero.
2. A long, insulating cylinder has a uniform charge density and a given radius.
a) To find the total charge on a specific length of the cylinder, you need to multiply the charge density by the volume of that length. The volume can be calculated by multiplying the cross-sectional area of the cylinder by its length.
b) To find the electric field both inside and outside of the cylinder, you can again apply Gauss' Law. Choose a Gaussian surface in the form of a cylindrical shell and calculate the total electric flux through that surface. Divide the total electric flux by the surface area of the Gaussian shell to find the electric field as a function of distance from the center of the cylinder.
c) To set up an integral to evaluate the potential at the surface of the cylinder, you need to consider the relationship between the potential and the electric field. Using the integral form of the potential, you can set up an integral that integrates the electric field over a path from the center to the surface of the cylinder.
d) When an isolated conducting shell is snapped around the insulating cylinder, its presence can affect the electric field at certain distances. To determine the effect on the electric field at r=50cm, you need to analyze how the charges redistribute due to the presence of the conducting shell. Consider the charges induced on the inner and outer surfaces of the conducting shell.
e) Similarly, to determine the effect on the potential at the surface of the insulator, you need to consider the redistribution of charges due to the presence of the conducting shell. Analyzing the charges induced on the inner and outer surfaces of the conducting shell will help you understand the qualitative effect on the potential.