help me solve this pls A metal sphere with radius ra = 1.30cm is supported on an

Question

help me solve this pls

A metal sphere with radius ra = 1.30cm is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb = 9.50cm. Charge + q is put on the inner sphere and charge - q on the outer spherical shell. The magnitude of q is chosen to make the potential difference between the spheres 490V , with the inner sphere at higher potential. Calculate q. Are the electric field lines and equipotential surfaces mutually perpendicular? Are the equipotential surfaces closer together when the magnitude of E rightarrow is largest? Equipotential surfaces closer together when the magnitude E rightarrow of is largest. Equipotential surfaces closer together when the magnitude E rightarrow of is smallest.

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) = 490 *4 * 3.1415 * 8.854e-12 / (1/0.013 - 1/0.095)

= 8.211E-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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