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The sketch below shows that two spherical perfect conductors are centered at the origin in the regions of R < b and c < R < d, respectively

Electrical Engineering

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The sketch below shows that two spherical perfect conductors are centered at the origin in the regions of R < b and c < R < d, respectively. The inner conductor, R - b, has a cavity of radius a, which is z d centered at (0, -yo,0), where 0 < yo < a < b/2 . The space between the two conductors is filled by a dielectric Dielectric b material of & = ErE,. There are no free charges in the system. Cavity 1.1. If positive charges of Q are placed on the inner a y surface of the inner conductor, use the knowledge of -yoo boundary conditions to determine: a. E at the conductor surfaces of z = b, c, and d. Conductor E b. E inside the cavity and the conductors. Hint: you need to find the surface charge densities first. Conductor 1.2. If the same amount of the charges Q as that from Problem 1.1 are placed on the inner surface of the outer conductor, determine the E at the locations described in Problem 1.1. 1.3. If the same amount of the charges Q as that from Problem 1.1 are placed on the center of the cavity, "picture" the electric field lines everywhere and determine E at the locations described in Problem 1.1. 1.4. Following Problem 1.3, refer to the sketch at right, P4 which of the following choices is (are) correct: P 3 a. V(P1)- V(P2) > 0 Dielectric b. V(P1)- V(R = b) > V(P1)- V(P2) c. V(P1)- V(P3) > V(P1)- V(P2) Cavity P 2 d. V(P])- V(R = d) > V(P1)- V(P3) e. V(P3)- V(R = d) > 0 -yoo f. V(P3)- V(R = c) = 0 g. V(P3)- V(PA) > V(R = d)- V(P4) Conductor h. V(P3) - V(P2) = V(R = c) - V(R = b) 2. Boundary Conditions: Conductor