Electric field outside of conducting sphere.

In summary, a conducting sphere with a charge of 1.0 × 10–9 C has a radius of 0.01 m. The electric field just outside the surface of the sphere can be calculated using the equation E=kq/r^2 or E=EA, where q is the charge, r is the radius, k is the Coulomb constant, and EA is the permittivity of the medium outside the sphere. After attempting the first equation, which resulted in a value of 90000, the poster was confused as the correct answer is 900. They asked for help and it was determined that the permittivity of the medium was not given in the question, leading to the incorrect calculation.
  • #1
ramses07
11
0

Homework Statement



A conducting sphere of radius 0.01 m has a charge of 1.0 × 10–9 C deposited on it. The magnitude of the electric field in N/C just outside the surface of the sphere is

Homework Equations


E=kq/r^2

EA=q/eo


The Attempt at a Solution



I tried the first equation and got 90000, but the answer is 900. I am racking my brains here trying to understand how to find the answer. Its seems simple enough, I am probably just missing a step, can anybody point to what I am doing wrong.
 
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  • #2
Could you show your calculations so far? Then can see where you're going wrong. :smile:
 
  • #3
Does the question mention the permittivity of the medium outside the sphere?
 
  • #4
..also I got [itex]E=90000NC^{-1}[/itex] as well. Where did you get that the answer should be [itex]E=900NC^{-1}[/itex]? Also check those values you've been given are correct (but they seem OK).
 
  • #5




It seems like you have the right equation, but you may have made a mistake in your calculation. The correct answer for the magnitude of the electric field just outside the surface of the conducting sphere is 900 N/C. To find this, you need to plug in the values of k (9 x 10^9 Nm^2/C^2), q (1.0 x 10^-9 C), and r (0.01 m) into the equation E = kq/r^2. This will give you a value of 900 N/C. It's possible that you may have made a calculation error, so double check your work to ensure you are getting the correct answer. Additionally, you could also use the second equation you listed, EA = q/eo, where eo is the permittivity of free space (8.85 x 10^-12 C^2/Nm^2). Plugging in the same values for q and r will also give you a result of 900 N/C. I hope this helps clarify the solution for you.
 

FAQ: Electric field outside of conducting sphere.

What is an electric field outside of a conducting sphere?

An electric field outside of a conducting sphere refers to the electric field that is present in the space surrounding a charged conducting sphere. This field is created by the electric charge on the surface of the sphere and can be measured using mathematical equations.

How does the electric field outside of a conducting sphere differ from the electric field inside the sphere?

The electric field outside of a conducting sphere is different from the electric field inside the sphere because the conducting material of the sphere allows the charges to move freely and redistribute themselves on the surface, resulting in a zero electric field inside the sphere.

What factors affect the strength of the electric field outside of a conducting sphere?

The strength of the electric field outside of a conducting sphere is affected by the magnitude of the charge on the sphere, the distance from the center of the sphere, and the dielectric constant of the surrounding medium. These factors can be mathematically calculated using Coulomb's law and Gauss's law.

Can the electric field outside of a conducting sphere be shielded or blocked?

Yes, the electric field outside of a conducting sphere can be shielded or blocked by surrounding the sphere with a material that is a good conductor. This material will redistribute the charges on its surface, canceling out the electric field from the sphere.

How is the electric field outside of a conducting sphere used in practical applications?

The electric field outside of a conducting sphere is used in various practical applications, such as in electrostatic painting, electrostatic precipitation, and electrostatic shielding. It is also important in understanding and analyzing the behavior of electrical charges and devices in the surrounding space.

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