An infinitely long solid insulating cylinder of radius a=3.2cm is positioned wit
ID: 3893222 • Letter: A
Question
An infinitely long solid insulating cylinder of radius a=3.2cm is positioned with its symmetry axis along the z-axis shown. The cylinder is uniformly charged with a charge density p=43mC/m^3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b= 12.5cm, outer radius c=15.5cm. The conducting shell has a linear density lambda = -.36mC/m.
1. What is Ey(R), the y-component of the electric field at point R, located a distance d =57cm from the origin along the y-axis as shown?
2.What is V(P)-V(R), the potential difference between the points P and R? Point P is located at (x,y) = (57cm, 57cm).
3.What is V(c)-V(a),the potential difference between the outer surface of the conductor and the outer surface of the insulator?
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?
5.The charge density of the insulating cylinder is now changed to a new value, ?
Explanation / Answer
1.
Q=p*a^2*piAn infinitely long solid insulating cylinder of radius a=3.2cm is positioned with its symmetry axis along the z-axis shown. The cylinder is uniformly charged with a charge density p=43mC/m^3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b= 12.5cm, outer radius c=15.5cm. The conducting shell has a linear density lambda = -.36mC/m.
then
E(R)=((lambda)+Q)/(r*2*pi*8.85e-12) , r= 0.57m
E(R)= (( -0.56e-6 C/m + 0.1061e-6 )/31.68e-12
E(R)= -1.4310x^4
2.
Use pythagorean theorem for point P to get your distance sqrt(.57^2+.57^2)
then = 0.8061
V(P)-V(R)= (((lambda)+Q)/(2*pi*8.85e-12))ln(P) - (((lambda)+Q)/(2*pi*8.85e-12))ln(R) ,where R= .57m
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. V(a) would be greater than zero because the there would be charge 0.56e-6 C/m accumulating on the outer surface of the insulating shell.
5. 0= (lambda+(rho*a^2*pi))/(2*pi*r*8.85e-12)
solve for rho