Calculate the magnitude and direction of the electric field

In summary, the spheres A and B have different radii and charges, and are conducting and concentric. The electric field and potential difference at various points between the spheres are calculated using the given equations. A small particle released from the outer surface of the inner sphere can also have its speed calculated using the conservation of energy principle.
  • #1
Shaad
13
0

Homework Statement



[1] Sphere A has a radius of 10.0 cm and carries a charge of +5.0 x 10^(-8) C
while sphere B has a radius of 5.0 cm and carries a charge of -2.0 x 10^(-8) C.
The spheres are conducting hollow, and concetric.

A) calculate the magnitude and direction of the electric field (E) at a point
3.00 cm from the center of the arrangement.


B) Calculate E at 6.00 cm from the center.


C) Calculate E at point 12.0 cm from the center.

D) Calculate the potential difference between the outer surface of the inner
sphere and the inner surface of the outer sphere.

E)A very small particle [ m = 2 x 10^(-9) kg, q = 1.5 x 10^(-9) is released from
the outer surface of the inner sphere. Calculate the speed with which it will
strike the inner surface of the outer sphere.


Homework Equations




E = q / eo = òE·dA [not sure I am using the right formula]

The Attempt at a Solution



I got an answer for a = 280
B) 70
C) 17.5
D) 15.7
E) no idea

But all of them are wrong. This problem is really irritating me, someone please help me out.

Homework Statement



[2] Two sheets od metal (A = 5.0 cm^2 are parallel and separated by 2.0 mm.
The potential difference between them is 1500V.
A) Calculate the electric field that exist between the plates.


B) Calculate the charge on the postive plate on the postive assuming the plates
have equal but opposite charges.

C) Calculate the work done in separating the plate to 3.0 mm apart.

Homework Equations





The Attempt at a Solution




No bloody idea about this problem.
 
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  • #2
This looks more like a physics problem than a math problem so I am going to move it to the physics homework section.

And I'm no physicist (as Doc Al has pointed out) but I was under the impression that the field inside a conducting sphere would be 0- so (a) would be 0. Similarly, the field in (b), where the point is between the two spheres, depends only on the smaller sphere.
 
  • #3
my bad i meant to post in physics section.

anyway thanks for the info. anyone else?
 
  • #4
B) Calculate E at 6.00 cm from the center.
What is E for a point charge of -2.0 x 10^(-8) C at distance 6.00 cm?

C) Calculate E at point 12.0 cm from the center.
What is E for a point charge of -2.0 x10^(-8)+ 5.0 x 10^(-8) C= 3.0 x 10^(-8)C at a distance of 12.0 cm?

D) Calculate the potential difference between the outer surface of the inner sphere and the inner surface of the outer sphere.
Just integrate.

E)A very small particle [ m = 2 x 10^(-9) kg, q = 1.5 x 10^(-9) is released from the outer surface of the inner sphere. Calculate the speed with which it will strike the inner surface of the outer sphere.
After you have done D this is just a matter of "conservation of energy". The decrease in potential energy is the increase in kinetic energy.
(Doc Al is now welcome to correct all my mistakes.)
 
Last edited by a moderator:

1. What is the formula for calculating the magnitude of the electric field?

The magnitude of the electric field is calculated by dividing the force on a charged particle by the magnitude of the charge on that particle. The formula is E = F/q, where E is the electric field, F is the force, and q is the magnitude of the charge.

2. How do you calculate the direction of the electric field?

The direction of the electric field is determined by the direction of the force on a positive test charge placed in the field. If the force is towards the source of the field, the direction of the electric field is towards the source. If the force is away from the source, the direction of the electric field is away from the source.

3. Can the magnitude of the electric field be negative?

Yes, the magnitude of the electric field can be negative. This occurs when the direction of the force on a charged particle is opposite to the direction of the electric field. However, the magnitude is always a positive value.

4. How does distance affect the magnitude of the electric field?

The magnitude of the electric field is inversely proportional to the distance between the charged particles. This means that as the distance increases, the magnitude of the electric field decreases. This relationship is described by the inverse square law.

5. Can the electric field have different magnitudes at different points?

Yes, the magnitude of the electric field can vary at different points in space. This is because the electric field is affected by the distance and distribution of charged particles in that area. The magnitude of the electric field is strongest closer to the source of the field and weaker further away.

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