An infinite line of charge with linear density 1 = -6.3C/m is positioned along t
ID: 1582449 • Letter: A
Question
An infinite line of charge with linear density 1 = -6.3C/m is positioned along the axis of a thick conducting shell of inner radius a = 2.9 cm and outer radius b = 4.7 cm and infinite length. The conducting shell is uniformly charged with a linear charge density 2 = 4.1 C/m.
A) What is Ex(P), the electric field at point P, located at (x,y) = (-11.3 cm, 0 cm) ?____N/C
B) What is Ey(P), the electric field at point P, located at (x,y) = (-11.3 cm, 0 cm)?______N/C
C) What is Ex(R), the electric field at point R, located a distance dR = 1.5 cm from the origin and making an angle of 30owith respect to the y-axis as shown?___N/C
D) What is Ey(R), the electric field at point R, located a distance dR = 1.5 cm from the origin and making an angle of 30owith respect to the y-axis as shown?_____N/C
E) What is b, the linear charge density on the outer surface of the conducting shell ?_____C/m
F) What is a, the linear charge density on the inner surface of the conducting shell ?_____C/m
G) The charged conducting shell is now replaced by an uncharged conducting shell of the same dimensions. What is b, the linear charge density on the outer surface of the uncharged conducting shell ?______C/m
H) The conducting shell is now given a new charge (2,new C/cm) such that the electric field at point P becomes equal to 0. What must be the sign of b, the linear charge density on the outer surface of the charged conducting shell?
b < 0
b = 0
b > 0
Explanation / Answer
Make sure to convert to proper units
Epsilon = 8.85x10^-12
1) E = (1 + 2) / (2pi)(r)(Epsilon)
=(-6.3+4.1)/2pi*2.9*8.85x10^-12
2) y direction so 0
3) E = ((1) / (2pi)(r)(Epsilon)) x cos(60)
-6.3/2pi*1.5*8.85x10^-12*cos(30)
4) E = ((1) / (2pi)(r)(Epsilon)) x sin(60)
=-6.3/2pi*1.5*8.85x10^-12*sin(30)
5) Just add 1 + 2 without converting
6) opposite charge 1 ( Make sure you check your signs on this one)
7) 1 no conversion
8) b = 0