A metal sphere with radius ra 1.40 cm is supported on an insulating stand at the

Question

A metal sphere with radius ra 1.40 cm is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb 9.70 cm. Charge +q is put on the inner sphere and chargeon the outer spherical shell. The magnitude of q is chosen to make the potential difference between the spheres 500 V,with the inner sphere at higher potential. Part A Calculate q Submit Request Answer . Part B Are the electric field lines and equipotential surfaces mutually perpendicular? O No Su Reauest Answer Part C Are the equipotential surfaces closer together when the magnitude of E is largest? O Equipotential surfaces closer together when the magnitude of E is largest. O Equipotential surfaces closer together when the magnitude of E is smallest. Provide Feedback Next >

Explanation / Answer

1) The potential difference of two concentric spheres is given by:

V = q / (4*pi*epsilon0) * (1/r1 - 1/r2)

resolving in q leads to:

q = V*4*pi*epsilon0 / (1/r1 - 1/r2) = 500 *4 * 3.1415 * 8.854e-12 / (1/0.014 - 1/0.094) = 9.10E-10 Coulomb

2) Yes, they are mutually perpendicular, because the E vector is the gradient of the electric potential. (Compare the electric potential to path of equal altitude around a mountain. The E vector can be compared to the direction and slope of the steepest descent (or ascent)).

3) a) Equipotential surfaces are closer together when the magnitude of E(vector) is largest.

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