1. Consider a plastic annulus of inner radius 3.90 cm and outer radius 12.00 cm.
ID: 2119107 • Letter: 1
Question
1. Consider a plastic annulus of inner radius 3.90 cm and outer radius 12.00 cm. The charge on the plastic annulus is uniformly distributed and amounts to a total of 65.0 %u03BCC.
(a) What is the charge density (a surface charge density) of the charged annulus?
1 C/m2
(b) For this problem, let the origin of the coordinate system be at the center of the annulus with the y and z directions in the plane of the annulus. Find the x-component of the electric field at the location (15.00 cm, 0, 0) due only to a tiny piece of the annulus of area 0.0267 mm2 which is located at y = 7.950 cm and z = 0.
2 N/C
(c) Find the electric field strength on the axis of the annulus at a distance of 15.00 cm from the center of the annulus.
3 N/C
(d) Find the electric field strength on the axis of the annulus at a distance of 30.00 cm from the center of the annulus.
4 N/C
(e) Imagine replacing the annulus by a point charge of 65.0 %u03BCC located at what had been the center of the annulus. What would then be the electric field strength at the same location used in parts (b) and (c)?
5 N/C
Explanation / Answer
(A)
Area of annulus = pi( (12)^2 - (3.9)^2 )*(10^ -4) = 0.04046
Charge = 65*(10^ -6) C
Charge density = Charge/ Area = 65*(10^ -6)/0.04046 = 1.6065*(10^ -3) C/m^2
(B)
Total electric field = k*q/r^2
Area = A =0.0267 mm^2 = 2.67*10^ -8 m^2
q = A*(Charge density) = (2.67*10^ -8)*(1.6065*10^ -3) = 4.289*(10^ -11) C
r = ( (15)^2 + (7.95)^2 )^1/2 = 16.97 cm = 0.17 m
Component along x - axis = [ k*q/r^2 ]*cos(w)
cos(w) = 15/16.97 = 0.8836
Put the values in [ k*q/r^2 ]*cos(w) and we get
Ex =[ (9*10^ 9)*4.289*(10^ -11) ] / (0.17)^2 * 0.8836 = 11.79 V/m
(C)
Electric field
d1 = ( (3.9)^2 + (15)^2 ) ^1/2 = 15.5 cm
d2 = ( (12)^2 + (15)^2 ) ^1/2 = 19.21 cm
Ex = 2*pi*k*(charge density) * [ x/d1 - x/d2 ] = 1.6976*(10^ 7) V/m
(D)
d1 = ( (3.9)^2 + (30)^2 ) ^1/2 = 30.25 cm
d2 = ( (12)^2 + (30)^2 ) ^1/2 = 32.31 cm
Ex = 2*pi*k*(charge density) * [ x/d1 - x/d2 ] = 5.74*(10^5) V/m
(E)
For part (b)
q = 65*10^ -6 C
r = 0.15 m
E = kq/r^2 = 2.6*10^7 V/m
For other part
q = 65*10^ -6 C
r = 0.3 m
E = kq/r^2 = 6.5*10^6 V/m
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