- #1
paul11273
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Here is a HW question that I have been wrestling with. Hopefully I have gone about solving it correctly. There is no answer in the back of the text to check my result against. Can anyone have a look and confirm?
Question-
"A solid plastic sphere with radius of 10.0cm has a charge with uniform density throughout it's volume. The electric field 5cm from the center is E=86.0 kN/C radially inward. Find the magnitude of the electric field 15cm from the center."
First I figured I needed to find the charge density. I set EA=qr/e0 ,where e0 is permittivity of free space and qr is total charge at radius of .05m given in the problem. I get to E=pr/3e0 where I am using p as charge density.
I solved for p with the given E and r, so p=4.6*10^-5 (C/m^3).
Next, I found the total charge within the .05m sphere with Q=pV. I got 1.91*10^-7 (C).
Lastly, I used E=ke(Q/r^2) to find the electrical field at radius=.15m. My final result is E=7.63*10^4 (N/C).
Can anyone tell me if I am correct? If not, where did I go wrong? Any help would be appreciated. I feel like I may have done too much work, there is probably a shorter way to the answer.
Thanks.
Question-
"A solid plastic sphere with radius of 10.0cm has a charge with uniform density throughout it's volume. The electric field 5cm from the center is E=86.0 kN/C radially inward. Find the magnitude of the electric field 15cm from the center."
First I figured I needed to find the charge density. I set EA=qr/e0 ,where e0 is permittivity of free space and qr is total charge at radius of .05m given in the problem. I get to E=pr/3e0 where I am using p as charge density.
I solved for p with the given E and r, so p=4.6*10^-5 (C/m^3).
Next, I found the total charge within the .05m sphere with Q=pV. I got 1.91*10^-7 (C).
Lastly, I used E=ke(Q/r^2) to find the electrical field at radius=.15m. My final result is E=7.63*10^4 (N/C).
Can anyone tell me if I am correct? If not, where did I go wrong? Any help would be appreciated. I feel like I may have done too much work, there is probably a shorter way to the answer.
Thanks.