Positive charge Q is distributed uniformly over each of two spherical volumes wi
ID: 1276115 • Letter: P
Question
Positive charge Q is distributed uniformly over each of two spherical volumes with radius R. One sphere of charge is centered at the origin and the other at x=2R
1. Find the magnitude of the net electric field due to these two distributions of charge at x=0 on the x-axis.
Express your answer in terms of the variables Q, R, and constants ? and ?0.
2. Find the magnitude of the net electric field due to these two distributions of charge at x=R/2 on the x-axis.
Express your answer in terms of the variables Q, R, and constants ? and ?0.
3. Find the magnitude of the net electric field due to these two distributions of charge at x=Ron the x-axis.
Express your answer in terms of the variables Q, R, and constants ? and ?0.
4. Find the magnitude of the net electric field due to these two distributions of charge at x=3Ron the x-axis.
Express your answer in terms of the variables Q, R, and constants ? and ?0.
Explanation / Answer
A. The charge of Q around the origin in sphere R will cancel out so there will be 0 effect from this one. The sphere of charge Q at 2R will act as a point charge of Q charge at 2R
E = (1/(4??o))(Q/(2R)^2)
B.
To get the net field at R/2 you can find the field due to each sphere, as if the other was not there, and add them as vectors.
The field inside the one sphere can be shown from Gauss' Law to be the field of a point charge, provided the charge used is only that amount beneath the field point.
So
E1 = kq/(R/2)^2 = 4kq/R^2 (k=1/4*pi*eo)
You can get "q" in terms of total "Q" by ratio & proportion;
q/(4/3)pi(R/2)^3 = Q/(4/3)piR^3
q = Q/8
Then
E1 = kQ/2R^2
The field due to the othe sphere is just due like the field of a point charge "Q" a distance (3/2)R away;
E2 = kQ/(3R/2)^2 = 4kQ/9R^2
At "R/2" on the axis these fields are in opposite directions so you subtract to get a net field.
Enet = E1 - E2 = kQ/18R^2 = Q/72*pi*eoR^2
C.
net electric field due to a point charge = k*q/r^2 where k = 1/(4*pi*e_0) where e_0 is the electric constant.
since at x=R, both spheres are touching, i.e. the electric field at the surface of the sphere, you could approximate the field at just outside the sphere. since both spheres' fields would be in equal and opposite directions, the net electric field
E_net = 0
D. A uniformly charged sphere would acts as a point charge placed at its center.
Hence the field due to the first sphere, E1=( 1/ 4? ?) Q1/ d1^2
And that due to the second sphere, E2= 1/ 4? ?) Q2/ d^2
As both the fields are in the positive direction of the x axis
the resultant field E = 9X10^9{ Q/ (9R^2 + Q/ (R^2)} = {9X10^9 X Q}/R^2{ 1/9 +1}
E = 9K*Q /(10*R^2)