A particle with a charge of -60.0 nC is placed at the center of a nonconducting
ID: 2119203 • Letter: A
Question
A particle with a charge of -60.0 nC is placed at the center of a nonconducting spherical shell of inner radius 19 cm and outer radius 31.0 cm. The spherical shell carries charge with a uniform density of -1.18 microC/m3.
(a) What is the electric field strength just barely inside the surface of the nonconducting shell?
N/C
(b) What is the total amount of charge on the nonconducting shell?
C
To answer the next three questions, it is useful to find an expression for the electric field strength inside the nonconducting shell as a function of distance from the center of the shell.
(c) What is the electric field strength at a distance of 22 cm from the center of the shell?
N/C
(d) What is the electric field strength at a distance of 25 cm from the center of the shell?
N/C
(e) What is the electric field strength at a distance of 28 cm from the center of the shell?
N/C
(f) What is the electric field strength just barely outside the surface of the nonconducting shell?
N/C
(g) Suppose a proton were to have a circular orbit just outside the spherical shell. Calculate the speed of this proton.
Explanation / Answer
a)
field due to shell inside shell is zero
hense
E = k*q1/r^2 = 9*10^9*-60*10^-9/0.19^2 = -14958.4487 N/m
b)
Volume of shell = 4/3*pi*(R2^3-R1^3) = 4/3*pi*(0.31^3-0.19^3)
charge on shell = d*V = -1.18*10^-6* 4/3*pi*(0.31^3-0.19^3) = -1.133*10^-7 C
c)
E inside spherical shell is given by formula
E = q/(4*eo*R)
q = charge placed at centre +charge due to shell
at R = 22 cm = 0.22 m
E = ((-60*10^-9)-(1.133*10^-7))/(4*8.854*10^-12*0.22) = -22242.1299 N/m
d)
at R = 25 cm = 0.25 m
E = (-60*10^-9-1.133*10^-7)/(4*8.854*10^-12*0.25) = -19573.0743 N/m
e)
at R = 28 cm = 0.28 m
E = (-60*10^-9-1.133*10^-7)/(4*8.854*10^-12*0.28) =-17475.9592 N/m
f)
E = k*(q)/r^2 = 9*10^9*((-60*10^-9)-(1.133*10^-7))/0.31^2 = -16229.9687 N/m
g)
F =Eq
so,
mV^2/r = E*q .......
V = sqrt(E*q*r/m)
V = sqrt(16229.9687*1.6*10^-19*0.31/(1.67*10^-27)) =694290.853 m/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.