Question

3. (8 points) Consider a conducting sphere with total electric charge +Q with radius Rị centered at p= 0 (spherical coordinates). The surface charge at r = R1 is spread uniformly on this spherical surface. There is also an outer conducting shell of radius r = R2, centered at r = 0 and with total electric charge - Q also spread uniformly on the surface. This arrangement of separated positive and negative charge forms a capacitor. We will assume that there is air between the inner and outer spheres. (a) i. Calculate the electric field Ē at a given radial position, r, between the 2 spheres (R1 < p < R2). ii. What is the electric Ē at a given radial position, r, outside the spheres (r > R2). iii. What is the electric E at a given radial position, r, in the inner sphere (r < Rı). (b) Calculate the potential difference from Rị to R2 given by Vrı – VR2. Perform this calculation using Ē calculated in part (a). (c) What is the capacitance, C = Q/V, of this capacitor? (d) Dielectric breakdown occurs when the electric field is so strong in a material that it causes the molecules in the material to rip apart. Describe briefly which region of the capacitor would be most likely to suffer dielectric breakdown when the electric field is strong.

Answer 1

Symmetrical charge distribution- 7) Consider a Gaussian sphere of radius r (R, <r <R) centered at the center of given capacitor. Electic field will be radially outwards on this Gaussian surface & since symmetry therefore magnitude of electric field will be some everywhere on this Gaussian surface. Hence Electric flux d=& E. dr = 6 E da caso" $ = Ef dA 6=E45592 - also by Gauss low, electric flux D = Lenclosed Eo . Euter = le radially from directed 2 bg, "R? 1 YTTER2 ľsky, Penelosed = +Q- Q = 0 = E= 0 as $ = 0

• (1) rar, again Venclosed to = E=0 (b) AVEC numerica E = levering OR du = Edr Ra και .AVE SËde = leve Δν dr had 0 AV = AV = e ate te or when - VA - Porte et ( cena ** d) E = RENT Lesser the 'r' more will be E and since R, <h, therefore E will be stronger near the conducting sphere (R1) hence region around ton immer conducting sphere (R.) would be most likely to suffer dielectric break down .