Magnitude of electric field near spherical shell

In summary: So you just need to add the field from the point charge to this. Can you write ##E## for the field from the point charge ?In summary, the problem involves a 10-nC point charge at the center of a thin spherical shell with a radius of 8.0 cm and a charge of -20 nC distributed uniformly over its surface. The question is asking for the magnitude of the electric field 2.0 cm from the point charge. After using the equation E = kq1q2/r^2, it was determined that the field from the shell can be ignored due to Gauss's theorem. Therefore, the field at 2.0 cm from the point charge is 0 N/C.
  • #1
jlmccart03
175
9

Homework Statement


A 10-nC point charge is located at the center of a thin spherical shell of radius 8.0 cm carrying -20 nC distributed uniformly over its surface.

What is the magnitude of the electric field 2.0 cm from the point charge?
upload_2017-2-8_13-23-15.png

Homework Equations


E = kq1q2/r^2

The Attempt at a Solution


I used k = 8.99*10^9 and the q charge with a distance of 0.08cm-0.02cm to get E = (8.99*10^9 C)(10*10^-9)(-20*10^-9)/(0.08-0.02)^2 = 9*10^5 N/C or 900 kN/C which is wrong. How do I go about doing this problem?
 
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  • #2
Your calculation is not in agreement with your relevant equation -- which is utterly wrong anyway (it calculates a force, not a field). Do you have a textbook with some relevant material you could study ?
 
  • #3
We don't use a textbook. Our professor has lecture notes instead.

EDIT Am I not even using the correct formula? I feel as though I am, but I may be WAY off. Which is bad considering I have an exam tomorrow over this.
 
Last edited:
  • #4
Relevant equation is ##E=k{q\over r^2}##

Did you learn about Gauss's theorem ? Tells you you can ignore the shell at 8 cm: it doesn't contribute to the field at 2 cm.
 
  • #5
BvU said:
Relevant equation is ##E=k{q\over r^2}##

Did you learn about Gauss's theorem ? Tells you you can ignore the shell at 8 cm: it doesn't contribute to the field at 2 cm.
OHHHH so then it should just be (8.99*10^9)(10*10^-9)/(0.02)^2 correct? And I still get confused on the uniform charge of sphere.
 
  • #6
jlmccart03 said:
OHHHH so then it should just be (8.99*10^9)(10*10^-9)/(0.02)^2 correct?
Correct if you add units.

jlmccart03 said:
And I still get confused on the uniform charge of sphere.
What is the field inside a uniformly charged sphere (just from this sphere)?
 
  • #7
mfb said:
Correct if you add units.

What is the field inside a uniformly charged sphere (just from this sphere)?
0 correct?
 
  • #8
Right.
 

Related to Magnitude of electric field near spherical shell

What is the formula for calculating the magnitude of electric field near a spherical shell?

The formula for calculating the magnitude of electric field near a spherical shell is E = kQ/r^2, where E is the electric field, k is the Coulomb's constant, Q is the charge on the shell, and r is the distance from the center of the shell.

How does the magnitude of electric field near a spherical shell vary with distance?

The magnitude of electric field near a spherical shell varies inversely with the square of the distance from the center of the shell. This means that as the distance increases, the electric field decreases.

What is the direction of the electric field near a spherical shell?

The direction of the electric field near a spherical shell is radially outward from the center of the shell. This means that the electric field lines point away from the shell in all directions.

Can the magnitude of electric field near a spherical shell be negative?

No, the magnitude of electric field near a spherical shell cannot be negative. The electric field is a vector quantity and its magnitude is always positive. However, the direction of the electric field can be negative if it points towards the center of the shell.

How does the magnitude of electric field near a spherical shell change with the charge on the shell?

The magnitude of electric field near a spherical shell is directly proportional to the charge on the shell. This means that as the charge on the shell increases, the electric field also increases. Similarly, if the charge on the shell decreases, the electric field also decreases.

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