Test Problems

  • Thread starter mmmboh
  • Start date
  • #1
407
0
Hi I have a test next week and am currently on spring break so I don't have a teacher to ask for help, I just want to confirm that some of these answers are correct and get some help for others.

1. A line with a charge of density -2uC is on the x-axis starting at point x=1m to x=3.5m. Find its charge density and derive an expression for the magnitude and direction of the electric field at a point situated at x=-1.5m.

Ok so here's what I did:

E=[tex]\int[/tex](KDQ)/r^2

I did [tex]\lambda[/tex]=dQ/dL and dQ=(Qdx)/2.5 (2.5 is the length of the line)

I took the limits to be 1 to 3.5, and r^2=2.5^2=6.25

So E=KQ/(2.5x6.25)[tex]\int[/tex]dx ...with limits 1 to 3.5

so E=KQ/15.6 (3.5-1), the final answer I got was 2.88x10^4 N/C, and the direction is left.

Is that right? if it isn't what do I change?

2. An insulating spherical shell with an inner radius 0.1 cm and outer radius 0.3 cm carries a total charge of 20 nC. Use Gauss's law to find an expression for the electric field at a distance r=0.08cm, r=0.2cm, r=0.4cm.

So what I did was [tex]\phi[/tex]=[tex]\int[/tex]EdA = Qin/E0.....integral of dA is A and A=4[tex]\Pi[/tex]r^2....
So then I did E=20nC/(4[tex]\Pi[/tex]r^2E0)...and I plugged in the different radii...however this is wrong, or most of it anyway, can someone help please?

3. An arc length with a length of 6 cm and a radius of 3cm carries a uniform charge of 10nC. Derive an expression for the magnitude and direction of electric field at the center.

I don't know what to do for this one. Am I suppose to integrate? like E=[tex]\int[/tex](KdQ)/r^2...

Thanks a lot for any help :)
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,097
5
For a point at -1.5 x, call it P, you have an integral that looks like k*dq/r2 = k*λ/r2*dx

where λ = c/l = 2 μC /2.5 m.
and r = (2.5 +x)

So don't you want to integrate k*λ/(2.5 + x)2*dx from 1 to 3.5?
 
  • #3
407
0
Ok so the integral is [tex]\int[/tex]K[tex]\lambda[/tex]dx/(2.5+x)^2 from 1 to 3.5.
Which becomes -K[tex]\lambda[/tex]/(2.5+x) from 1 to 3.5. And so the final answer is 8.57 x 10^9 C/m to the left?

But why is r (2.5 + x)?....is it because x=-1.5 and the start of the line is at 1, and the difference is 2.5...and then x is just a place where it could be on the line? I am confused :S
 
  • #4
LowlyPion
Homework Helper
3,097
5
Ok so the integral is [tex]\int[/tex]K[tex]\lambda[/tex]dx/(2.5+x)^2 from 1 to 3.5.
Which becomes -K[tex]\lambda[/tex]/(2.5+x) from 1 to 3.5. And so the final answer is 8.57 x 10^9 C/m to the left?

But why is r (2.5 + x)?....is it because x=-1.5 and the start of the line is at 1, and the difference is 2.5...and then x is just a place where it could be on the line? I am confused :S

That's right. r is the displacement function from the point that you are measuring to the closest dq that you are going to integrate. So in the equation for the E field you have

dE = k*λ/r2*dx

But since r is a function of x, as in r(x) = (2.5 + x)
 
  • #5
407
0
Thanks :)...how bout this one:

Two perpendicular conducting sheets with area 2m^2 each. One has a charge of +10uC and the other +30uC uniformly spread over their surfaces. Find the magnitude of the electric field at the point (0.4m, 0.5m) from the corner. Neglect the end effects.

So what I did was E=[tex]\sigma/2E0 [/tex] (Sorry I couldn't find the absolute not symbol)

So then I found [tex]\sigma1[/tex] = 15uC/m^2, and [tex]\sigma2[/tex]=5uc/m^2, and I plugged them both into the equation E=[tex]\sigma/2E0 [/tex] and then I squared both of them and added them together, and then I square rooted it to find the magnitude...and I neglected the points (0.4m, 0.5m)...the final answer I got is 9.18x10^4 N/C. Did I do this properly?

I have a test Monday, and am on spring break so I have no one to ask, help is much appreciated :)
 
  • #6
LowlyPion
Homework Helper
3,097
5
Looks like a good method.

Your squares are 1.4m on a side, so the .4,.5 coordinate looks like you can rely on the construction that the E field from both components can be given by:

E = σ/2εo

Taking the RSS of 5 and 15, in x and y, I get 15.5μC/2εo
 
  • #7
407
0
Sorry I don't understand. The square is 1.4m on both sides? are you talking about a gaussian surface, if so why 1.4?...and what does RSS mean?
 
  • #8
LowlyPion
Homework Helper
3,097
5
Sorry I don't understand. The square is 1.4m on both sides? are you talking about a gaussian surface, if so why 1.4?...and what does RSS mean?

2m2 means 1.4 m on a side.

Root Sum of the Squares is shortened to RSS.
 
  • #9
407
0
Ok I did:
[tex]\sigma1[/tex]=10uC/2m^2=5uC/m^2
and [tex]\sigma2[/tex]=30uC/2m^2=15uC/m^2

and then I plugged them in to the equation E=[tex]sigma/(2εo)[/tex]

So E1 I got is 2.83x10^11 and E2 is 8.48x10^11

And then I RSS'd them and got 8.94x10^11 N/C....it's a little bit different from yours, but I think it's right?
 
  • #10
LowlyPion
Homework Helper
3,097
5
Ok I did:
[tex]\sigma1[/tex]=10uC/2m^2=5uC/m^2
and [tex]\sigma2[/tex]=30uC/2m^2=15uC/m^2

and then I plugged them in to the equation E=[tex]sigma/(2εo)[/tex]

So E1 I got is 2.83x10^11 and E2 is 8.48x10^11

And then I RSS'd them and got 8.94x10^11 N/C....it's a little bit different from yours, but I think it's right?

Probably, because mine was in error, I typo'd 15.5 when I meant 15.8.

But it look's like a factor of μ is missing in your final magnitude.
 
  • #11
407
0
Ah, thanks a lot for the help! :)
 

Related Threads on Test Problems

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
13
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
870
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
4
Views
1K
Top