An infinitely long solid insulating cylinder of radius a = 5.9 cm is positioned
ID: 1532439 • Letter: A
Question
An infinitely long solid insulating cylinder of radius a = 5.9 cm is positioned with its symmetry axis along the z-axis as shown. The cylinder is uniformly charged with a charge density = 46 C/m3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 18 cm, and outer radius c = 20 cm. The conducting shell has a linear charge density = -0.59C/m.
1)
What is Ey(R), the y-component of the electric field at point R, located a distance d = 44 cm from the origin along the y-axis as shown?
N/C
2)
What is V(P) – V(R), the potential difference between points P and R? Point P is located at (x,y) = (44 cm, 44 cm).
V
3)
What is V(c) - V(a), the potentital difference between the outer surface of the conductor and the outer surface of the insulator?
V
4)
Defining the zero of potential to be along the z-axis (x = y = 0), what is the sign of the potential at the surface of the insulator?
V(a) < 0
V(a) = 0
V(a) > 0
5)
The charge density of the insulating cylinder is now changed to a new value, ’ and it is found that the electric field at point P is now zero. What is the value of ’?
µC/m3
Explanation / Answer
First you have to change p(rho) into a charge. by times it by the area
1.
Q=p*a^2*pi
then
E(R)=((lambda)+Q)/(r*2*pi*8.85e-12) , r= .44m
2.
Use pythagorean theorem for point P to get your distance sqrt(.44^2+.44^2)
then
V(P)-V(R)= (((lambda)+Q)/(2*pi*8.85e-12))ln(P) - (((lambda)+Q)/(2*pi*8.85e-12))ln(R) ,where R= .44m
3. ok so you have to take the integral from a to b of just the inner cylinder.
-(Q/(2*pi*8.85e-12))integral(a to b) of (1/r)dr
4. 0= (lambda+(rho*a^2*pi))/(2*pi*r*8.85e-12)