Please show step by step solution to answer for the magnitude and direction of t
ID: 1527585 • Letter: P
Question
Please show step by step solution to answer for the magnitude and direction of the electric field, thank you ! :)
A conducting spherical shell of inner radius a 50.0 cm and outer radius b- 60.0 cm has a net charge Q1 7.00 HC. A second large r conducting shell of inner radius 70.0 cm and outer radius d 80.0 cm has a net charge of Q2 3.00 HC, and it is concentric with the first shell as shown in the figure to the left. What is the magnitude and direction of the electric field as a function of distance from the center of the spherical shells?Explanation / Answer
as per gauss law,
total electric flux passing through a surface is equal to charge enclosed/ electrical permitivity
let electric field be E .
we will consider different intervals with concentric circle of radius r.
for r<a:
charge enclosed =0
so electric field =0
for a<r<b:
charge enclosed by a surface of radius r=Q1*pi*(r^2-a^2)/(pi*(b^2-a^2))
=Q1*(r^2-0.5^2)/(0.6^2-0.5^2)=-0.77*(r^2-0.25) uC
if electric field is E,
then epsilon*E*4*pi*r^2=charge enclosed
==>E=-0.77*(r^2-0.25)*10^(-6)/(4*pi*r^2*8.85*10^(-12))=- 6923.6896*(r^2-0.25)/r^2 N/C
for b<r<c:
charge enclosed=Q1=-7 uC
so if electric field is E,
8.85*10^(-12)*E*4*pi*r^2=-7*10^(-6)
==>E=-62942.63286/r^2 N/C
for c<r<d:
charge enclosed=Q1+(Q2*(r^2-c^2)/(d^2-c^2))
=-7+(3*(r^2-0.7^2)/(0.8^2-0.7^2))=-7+20*(r^2-0.49) uC=20*r^2-16.8 uC
if electric field is E,
then 8.85*10^(-12)*E*4*pi*r^2=(20*r^2-16.8)*10^(-6)
==>E=(20*r^2-16.8)*8991.8/r^2 N/C
r>d:
charge enclosed=Q1+Q2=-4 uC
then E=-4*10^(-6)*9*10^9/r^2
=-36000/r^2 N/C
direction of electric field:
when ever electric field is negative, field is directed radially inwards.
when the field is positive, the field is directed radially outward.