point charge q 1 = -7.6 C is located at the center of a thick conducting shell o
ID: 1528287 • Letter: P
Question
point charge q1 = -7.6 C is located at the center of a thick conducting shell of inner radius a = 2.2 cm and outer radius b = 4.1 cm, The conducting shell has a net charge of q2 = 1.7 C.
1)
What is Ex(P), the value of the x-component of the electric field at point P, located a distance 6.1 cm along the x-axis from q1?
N/C
2)
What is Ey(P), the value of the y-component of the electric field at point P, located a distance 6.1 cm along the x-axis from q1?
N/C
3)
What is Ex(R), the value of the x-component of the electric field at point R, located a distance 1.1 cm along the y-axis from q1?
N/C
4)
What is Ey(R), the value of the y-component of the electric field at point R, located a distance 1.1 cm along the y-axis from q1?
N/C
5)
What is b, the surface charge density at the outer edge of the shell?
C/m2
6)
What is a, the surface charge density at the inner edge of the shell?
C/m2
7)
For how many values of x: (4.1 cm < x < infinity) is it true that Ex = 0?
none
one
more than one
8)
Define E2 to be equal to the magnitude of the electic field at r = 1.1 cm when the charge on the outer shell (q2) is equal to 1.7 C. Define Eo to be equal to the magnitude of the electric field at r = 1.1 cm if the charge on the outer shell (q2) were changed to 0. Compare E2 and Eo.
E2 < Eo
E2 = Eo
E2 > Eo
i q RxExplanation / Answer
Epsilon = 8.85x10^-12
1) At point P both the conducting shell and the inner point charge are enclosed and we add both together to get Qenclosed.
(Qenclosed)/(Epsilon)(4pi)(p^2) = (9*10^9*5.9*10^-6)/(0.061)^2 = -1.43*10^7 N/C
2) Since Y is perpendicular to the radial axis here, none of the E field will be in the Y direction
ZERO
3) Here the X axis is perpendicular to the radial direction and so there is no E field component in that direction.
ZERO
4) The only differences are that the radius is shorter than in (1) and that only the point charge is enclosed now.
E = (q1) / (Epsilon)(4pi)(R^2) = (9*10^9*7.6*10^-6)/(0.011)^2 = -5.65*10^8 N/C
5) (q1 + q2) / (4pi)(b^2) = (-7.6 + 1.7)*10^-6/(4pi*0.041^2) = -2.8*10^-4 C/m^2
6) We know for a conductor the interior has an E field of zero. Since we saw that the E field is proportional to the charge enclosed, to get E=0 we need Qenclosed=0. That must mean that Qpoint+Qinnner shell=0 or in other words the charge on the inner shell must be exactly equal and opposite to the point charge. Then just divide by the area of a sphere to get charge density.
= (q1) / (4pi)(a^2) = 1.25*10^-3 C/m^2
7) None
The field is treated as if it is a single point charge outside the conducting wall and there after extends to infinity diminishing by a rate of [r^2].
8) E2 = E0
The fields are equal as the charge on the outer shell does not effect the field on within the shell.