Electric Fields and point charges

[beachy6]

Homework Statement

A point charge −3 pC is concentric with two
spherical conducting thick shells, as show in
the figure below. The smaller spherical conducting shell has a net charge of 22 pC and
the larger spherical conducting shell has a net
charge of 18 pC.

1-4)Answer in units of pC
What is the charge on the inner surface of
the smaller spherical conducting shell, 1.2 cm
from the −3 pC point charge? Under static
conditions, the charge on a conductor resides
on the surface of the conductor.
Answer in units of pC


What is the charge on the outer surface of
the smaller spherical conducting shell, 4.1 cm
from the −3 pC point charge?

What is the charge on the inner surface of the
larger spherical conducting shell, 6 cm from
the −3 pC point charge?

What is the charge on the outer surface of the
larger spherical conducting shell, 8.8 cm from
the −3 pC point charge?

5-9)Answer in units of N/C
What is the magnitude of the electric field at
P1, 0.6 cm from the −3 pC point charge?

What is the magnitude of the electric field at
P2, 2.65 cm from the −3 pC point charge?

What is the magnitude of the electric field at
P3, 5.05 cm from the −3 pC point charge?

What is the magnitude of the electric field at
P4, 7.4 cm from the −3 pC point charge?

What is the magnitude of the electric field at
P5, 10.2 cm from the −3 pC point charge?





2. Homework Equations
electric flux= in E*dA
E=(kq/r^2)

3. The Attempt at a Solution
I started solving these and I know I have to use Gauss Law, I know I need to apply Gauss law for different radii but when i try set it up electric flux= in E*dA I don't seem to get it. I have tried several different ways

I know that 6 and 8 are 0 because they are in the conductor.
7 i used E=(kq/r^2)

but i have been really confused on the rest of them.

I have tried many different equations but have been confused on different lengths! If someone could at least lead me in the right direction I would really appreciate it!

 

[ehild]

Make a drawing. What is the electric field inside the shells? Show the surface charge densities on the shells, and show your work in detail, please.

ehild

 

[beachy6]

Here is a picture. I drew it myself so it may not be the best drawn picture, but it is accurate to the problem. Thanks a lot for the help!

 

[ehild]

Nice picture!

You know that the charges of a metal accumulate on the surface, and the electric field is zero inside the metal.
Also the number of electric field lines crossing a unit surface normally is equal to the surface charge density divided by ε₀.

So what is the charge on the inner surface of the smaller shell?

ehild

 

[beachy6]

Wouldnt the charge of the inner surface of the smaller shell simply E*dA, since the radius is 1.2 cm (0.012m) would it be

E*dA= (-3pC)*(pi*0.012^2)?

 

[ehild]

εI do not understand what you wrote. The whole charge on the inner surface is εEA.
What is the electric field at 0.012 m from the centre?


ehild

 

[beachy6]

Would you do take the area and multiply it by the charge? so (pi)(r^2)(-3pC)?

 

[mmiller12]

hmmmm i tried using E=q/SA, which was not correct either.
Not exactly sure where to go from here then

sorry bud

 

[ehild]

The electric field has to be multiplied by ε₀ and the surface area of the sphere. -3 pC is a point charge. What is the electric field at a distance r=0.012 m from it, according to Coulomb's law?

ehild

 

[beachy6]

Electric field would be equal to the point charge/ (surface area)(ε0) correct?

 

[beachy6]

So for number one when it asks "What is the charge on the inner surface of
the smaller spherical conducting shell, 1.2 cm
from the −3 pC point charge? would i use the equation
q=(electric field)(surface area)(E0)

but how do i fiend the electric field?

 

[beachy6]

This problem is so frustrating! I have tried everything possible i feel like. Anyone know if for numbers 7 and 9 we have to take into account the conuctors and subtract the radius by the width of the conductor?

 

[ehild]

Answer question 1 first. Show what you tried.

ehild

 

[ehild]

So for number one when it asks "What is the charge on the inner surface of
the smaller spherical conducting shell, 1.2 cm
from the −3 pC point charge? would i use the equation
q=(electric field)(surface area)(E0)

but how do i fiend the electric field?

Look at your previous post.

Electric field would be equal to the point charge/ ((surface area)(ε0))

Take the direction of the electric field into account. What kind of charge does accumulate on the inner surface, positive or negative?
ehild

 

[beachy6]

Since it is in the inner surface it would be a negative because the formula is -q/4pir^2Once i find the electric field, how do i use this to find the charge? we have tried E*Area=Q/E0

 

[mmiller12]

This is a really tough problem. You might need to take a question like this to a professor or someone at your school. I usually try to get help from a professor during office hours. A problem like this might be too difficult for you to get help from people on physics forum. It is very involved. Hope you have good luck on this site though,

 

[ehild]

We try on the other way. There is a negative point charge in the middle. The electrons of the metal shell are free to move. Are they attracted or repulsed by the -3 pC charge?

ehild

 

[beachy6]

Ya i have asked several of my friends who have taken this class for help but they all could not remember exactly how to do a problem like this. My professor has really bad office hours and is not too helpful... thanks tho for the advice

 

[beachy6]

repulsed

 

[ehild]

OK. So what kind of charges remain on the inner surface if the negative charges are repulsed away?

ehild

 

[beachy6]

A positive charge. So if we have a positive charge would the answer to #1 just be a +7pC?

 

[beachy6]

Sorry my teacher did not explain this well at all, and expects us to know things that he doesn't discuss in class. thanks for helping i appreciate it!

 

[ehild]

Ok there will be positive charge on the inner surface, but why 7 pC?
Do you know Gauss' Law? The electric field multiplied with the area of a closed surface is equal to the enclosed charge divided by ε₀. Take a sphere inside the metal shell.E is zero, so the total enclosed charge is zero. Enclosed is the charge in the middle and the charge on the inner surface. So how much charge is on the inner surface?

ehild

 

[beachy6]

When I use that Equation and use 7 as my charge I get the Electric field i get 1.09226 x 10^14 pC/N.

But if I use 0, the obviously the electric field would be zero, but i don't think it can be zero since it is not in the conductor

 

[ehild]

Where is that 7 pC from?

ehild

 

[beachy6]

Oops I am Sorry, I was looking at the picture wrong.What I meat to say is use the 3pC from the center charge and plug that into the equation and get 4.68x10^13 N/pC

 

[ehild]

The charge in the centre is negative, so the electric field points toward the centre. There are positive charges on the inner surface of the shell. The electric field lines originate from positive charges and terminate in the negative ones. There must be equal and opposite charges at both ends of an electric field line. So how much charge is accumulated on the inner surface of the small shell?

ehild

 

[ehild]

Show how you calculate the electric field of the central -3pC charge at distance r=0.012 m.

ehild

 

[beachy6]

The electric field= 3pC/(ε0*SA)

 

[ehild]

Could you please evaluate?

ehild

 

[beachy6]

ya, E=3pC/ (8.854x10^-12)(4pi (.012^2))= 4.68x10^13 N/pC

 

[beachy6]

So How do i use this value to now find the charge on the inner surface?

 

[ehild]

ya, E=3pC/ (8.854x10^-12)(4pi (.012^2))= 4.68x10^13 N/pC

It is wrong. How did you get that number?

ehild

 

[beachy6]

O i got it, so it would be +3pC