Shows two concentric, thin conducting spherical shells with radii r_a & r_b resp

Question

Shows two concentric, thin conducting spherical shells with radii r_a & r_b respectively (r_a > r_b). The inner shell (radius r_b) carries a uniformly distributed net positive charge Q_b and the outer shell (radius r_a) carries a uniformly distributed net negative charge Q_a. The cavity inside the inner shell (r_b > r) and the space between the shells (r_a > r > r_b) are empty. For the following, please remember that, as was just clearly stated, the charges on the inner and outer shells are NOT equal and opposite Calculate: The electric flux Phi_E passing through a spherical Gaussian surface of radius r inside the inner shell and the electric field E everywhere inside the inner shell (r_b > r). The electric flux Phi_E passing through a spherical Gaussian surface of radius r between the two spherical shells and the electric field E everywhere between the shells (r_a > r > r_b). The electric flux Phi_E passing through a spherical Gaussian surface of radius r outside the outer shell and the electric field E everywhere outside the outer shell (r > r_a). The electric potential difference delta V = V_b - V_a between the two metal spheres. To obtain the numerical answers asked in the following, use r_a = 0.2 m, r_b = 0.4 m Q_b = 3.0 Times 10^-5 C, and Q_a = - 1.5 Times 10^-5 C. Evaluate the electric field numerically for r = 0.3 m and for r = 0.5 m. A point charge q = 5.0 Times 10^-6 C is placed a distance r = 0.6 m from the center of the spheres. Numerically evaluate the magnitude F of the force it experiences there. What is the vector direction of this force? The following may be needed: Differential volume element in spherical coordinates (with no angular dependence): dV = 4pi^r^2dr. Integral of r to a power Integral r^ndr = [r^n+1/(n+1)J (n -1).

Explanation / Answer

a) Flux = QEnclosed /eo

= 0 because Qenclosed = 0

E.4pi r^2 = 0

E = 0

B) Flux = QEnclosed /eo

   = Qb / eo because Qenclosed = 0

E.4pi r^2 = Qb / eo

E = Qb / ( 4 pi eo r^2)

c) Flux = QEnclosed /eo

   = (Qb + Qa) /eo because Qenclosed = 0

E.4pi r^2 = (Qb + Qa) / eo

E = (Qb+ Qa) / ( 4 pi eo r^2)

d) delta V = Vb- Va

   = Qb */ ( 4 pi eo rb) - (Qb + Qa) / ( 4 pi eo ra)

= 9e+9 * 3e-5/0.2 - 9e+9*1.5e-5/0.4

= 1012500 V

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