(a) How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 30.0 cm in diameter to produce an electric field of 1150 N/C just outside the surface of the sphere?
(b) what is the electric field at a point 10.0 cm outside the surface?
ok would I use Guass's law for this one? I am lost... and with the 10 cm..where do I put that in the problem?
a) Use Gauss' law.
4 pi R^2 E = N*e/(epsilono)
N is the number of electrons and e is the electron charge in Coulombs.
E is the electric field strangth in N/C
"epsilon zero" = 8.8554*10^-12 C^2/N m^2
Make sure R is in meters
b) Use the same equation, but use R + 0.1 m in place of R since you are 10 cm farther away.
ok..thank you very much. I thought that is what you would do for part b but I wasn't sure! Thank you!
Well, Gauss's law would certainly be useful here. Don't worry, I'll help you through it!
For part (a), we need to find the total charge inside the sphere. We know that the electric field just outside the surface is 1150 N/C. Using Gauss's law, we can relate the electric field to the charge enclosed as:
E = Q / (4πε₀r²),
where E is the electric field, Q is the charge enclosed, ε₀ is the permittivity of free space, and r is the radius of the sphere.
Since the sphere is isolated, the charge is evenly distributed throughout the volume. The volume of a sphere is given by V = (4/3)πr³, so the total charge Q can be calculated as Q = n * e, where n is the number of excess electrons and e is the charge of an electron.
Now, the radius of the sphere is given as 30.0 cm, so the electric field just outside the surface is E = 1150 N/C. Plugging in these values into the equation, you can solve for Q.
For part (b), we need to find the electric field at a point 10.0 cm outside the surface of the sphere. We can use the same formula as before, but now we need to use the new distance r = 10.0 cm (the distance from the center of the sphere to the point where we want to measure the electric field).
With that, you can plug in the values into the equation and solve for the electric field at that point.
I hope this helps! Let me know if you have any more questions or need further assistance.
To determine the number of excess electrons required to produce a specific electric field just outside the surface of the plastic sphere, you can use Coulomb's law and the concept of electric field strength.
Here's how you can approach the problem:
(a) To find the number of excess electrons, you'll need to start with the equation for the electric field due to a point charge:
E = k * (q / r^2)
Where:
- E is the electric field strength just outside the surface of the sphere
- k is the electrostatic constant (k ≈ 8.99 x 10^9 Nm^2/C^2)
- q is the charge of an electron (q ≈ 1.6 x 10^-19 C)
- r is the distance from the center of the sphere to the point just outside the surface
Since the sphere is electrically neutral, the total excess charge is equal to the total charge of the excess electrons. Let N be the number of excess electrons. Each electron has a charge of -e, so the total charge is:
q_total = N * -e
Given that the diameter of the sphere is 30.0 cm, the radius (r) is 15.0 cm (or 0.15 m).
To find the number of excess electrons, we need to solve the equation E = k * (q_total / r^2) for N:
N * -e = E * r^2 / k
N = -E * r^2 / (e * k)
Substituting the given values:
E = 1150 N/C
r = 0.15 m
e ≈ 1.6 x 10^-19 C
k ≈ 8.99 x 10^9 Nm^2/C^2
Plug in the values and calculate N to find the number of excess electrons.
(b) To find the electric field at a point 10.0 cm outside the surface, you'll use the same equation:
E = k * (q / r^2)
This time, however, r will be the distance from the center of the sphere to the point, rather than the radius of the sphere. So, for this case, r = 10.0 cm = 0.10 m. Plug in the given values in the equation and calculate the electric field.
Remember to convert all distances into meters and use the appropriate values for the constants in the calculations.