Calculating the charge if the electric field density = 0

In summary, r is the radius of the sphere, a is the radius of the sphere's center, dv is the distance between the center and the sphere's surface, dr is the distance between the center and the sphere's center of mass, and ρ is the charge at the center of the sphere.
  • #1
falyusuf
35
3
Homework Statement
Attached below.
Relevant Equations
Attached below.
Question:
846%2F8468799a-80b2-4052-85ae-161b13a4fffa%2Fimage.jpg

Relevant Equations:

1637429860368.png

1637429882368.png

My attempt:
1637429910033.png

1637429929135.png

Could someone please confirm my solution?
 
Physics news on Phys.org
  • #2
Doesn't look right.
1. Is r a variable or a constant?
2. Is dv = drdθdφ?
3. Is the integral of r3/100 = a3/100?
 
  • #3
mjc123 said:
1. Is r a variable or a constant?
r is constant as it is the radius of the sphere whereas a is variable.
mjc123 said:
2. Is dv = drdθdφ?
yes, (for the sphere)
mjc123 said:
3. Is the integral of r3/100 = a3/100?
I was mistaken here, it should be a^4 / 400
 
  • #4
falyusuf said:
r is constant as it is the radius of the sphere whereas a is variable.
You are integrating by dr from limits 0 to a, yet treat 4πr2 as a constant.
falyusuf said:
yes, (for the sphere)
Really? What are the dimensions on either side?
 
  • #5
mjc123 said:
You are integrating by dr from limits 0 to a, yet treat 4πr2 as a constant.

Really? What are the dimensions on either side?
Is dv = r^2 sin theta dtheta dPhi dr ?
I tried to correct my mistakes and this is what I got:
1637438382719.png
 
Last edited:
  • #6
I still think you're using r in a confusing way. Use r for the integration variable and R (constant) for the distance at which you want to evaluate the field. Then you are not integrating r3, but r5/R2.
 
  • #7
mjc123 said:
I still think you're using r in a confusing way. Use r for the integration variable and R (constant) for the distance at which you want to evaluate the field. Then you are not integrating r3, but r5/R2.
Am I right now?
1637445069498.png
 
  • #8
The integration variable is r, not R. And the upper limit is a, not R.
 
  • #9
mjc123 said:
The integration variable is r, not R. And the upper limit is a, not R.
Sorry I was confused.
1637461115389.png

1637461154289.png


Right?
 

Attachments

  • 1637446546109.png
    1637446546109.png
    7.8 KB · Views: 108
Last edited:
  • #10
R is not 0.1 m. a is 0.1 m. You should have R2 in the denominator for both expressions, and it will cancel out.
You need to be very careful in distinguishing the various distances and using consistent notation for them. Too many rs is a recipe for confusion.
 
  • #11
mjc123 said:
R is not 0.1 m. a is 0.1 m. You should have R2 in the denominator for both expressions, and it will cancel out.
You need to be very careful in distinguishing the various distances and using consistent notation for them. Too many rs is a recipe for confusion.
Thanks for clarifying. I solved it using two methods and I got the same magnitude with opposite sign. Could you please figure out my mistake?
1637589653658.png

1637589697386.png
 
  • #12
I'm not quite sure what you're doing in method 2, but the point charge at the centre should be equal to minus the charge obtained by integrating ρ over the volume of the sphere. So from outside it behaves like a point charge of zero.
I think your answer of -2.09e-8 C is correct. (I did it in my head and got -2.09e-4 C; I think I must have accidentally switched the factor of 100 from the bottom to the top:).)
 
  • Like
Likes falyusuf
  • #13
mjc123 said:
I'm not quite sure what you're doing in method 2, but the point charge at the centre should be equal to minus the charge obtained by integrating ρ over the volume of the sphere. So from outside it behaves like a point charge of zero.
I think your answer of -2.09e-8 C is correct. (I did it in my head and got -2.09e-4 C; I think I must have accidentally switched the factor of 100 from the bottom to the top:).)
I got it now.. Thank you so much. Appreciate your help.
 
Last edited:

FAQ: Calculating the charge if the electric field density = 0

What is the formula for calculating the charge if the electric field density is 0?

The formula for calculating charge when the electric field density is 0 is Q = 0, where Q represents the charge.

How is the electric field density related to charge?

The electric field density is directly proportional to charge. This means that as the charge increases, the electric field density also increases.

Can the charge be calculated if the electric field density is 0?

Yes, the charge can still be calculated even if the electric field density is 0. This is because the formula Q = 0 will still yield a value of 0 for the charge.

What are some real-life examples of situations where the electric field density is 0?

One example is inside a hollow metal sphere, where the electric field is 0 at all points inside the sphere. Another example is at the center of a uniformly charged spherical shell, where the electric field is 0 due to symmetry.

How does the electric field density affect the behavior of charged particles?

The electric field density determines the strength and direction of the electric field, which in turn affects the motion and behavior of charged particles. For example, a higher electric field density will result in a stronger force on a charged particle, causing it to accelerate more quickly.

Similar threads

Replies
2
Views
1K
Replies
7
Views
2K
Replies
11
Views
2K
Replies
3
Views
2K
Replies
11
Views
2K
Replies
26
Views
1K
Replies
1
Views
1K
Back
Top