1) What is E y (R), the y-component of the electric field at point R, located a

Question

1) What is Ey(R), the y-component of the electric field at point R, located a distance d = 52 cm from the origin along the y-axis as shown?

2) What is V(P) – V(R), the potential difference between points P and R? Point P is located at (x,y) = (52 cm, 52 cm).

3) What is V(c) - V(a), the potentital difference between the outer surface of the conductor and the outer surface of the insulator?

4) 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 ?’?

An infinitely long solid insulating cylinder of radius a = 2.7 cm is positioned with its symmetry axis along the z-axis as shown. The cylinder is uniformly charged with a charge density = 30 HC/m3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 15.4 cm, and outer radius c = 19.4 cm. The conducting shell has a linear charge density A- -0.31uC/m. R(0.d) Pad) ar charge density =

Explanation / Answer

so 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= .52m

2.
Use pythagorean theorem for point P to get your distance sqrt(.52^2+.52^2)
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= .52m

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)
solve for rho

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