For a single, isolated point charge carrying a charge of q = 74.2 pC, one equipo

Question

For a single, isolated point charge carrying a charge of q = 74.2 pC, one equipotential surface consists of a sphere of radius 32.9 mm centered on the point charge as shown. What is the potential on this surface? You would like to draw an additional equipotential surface, which is separated by 7.82 V from the previously mentioned surface. How far from the point charge should this surface be? This surface must also meet the condition of being farther from the point charge than the original equipotential surface.

Explanation / Answer

(a) The potential on this surface which is given by :

using an equation, we have

V = ke q / r

where, q = point charge = 74.2 x 10-12 C

r = distance from the center = 32.9 x 10-3 m

V = (9 x 109 Nm2/C2) (74.2 x 10-12 C) / (32.9 x 10-3 m)

V = (667.8 x 10-3 Nm2/C) / (32.9 x 10-3 m)

V = 20.3 V

(b) The distance from the point charge which will be given as :

from an above equation, we get

V = ke q / r

(7.82 V) = (9 x 109 Nm2/C2) (74.2 x 10-12 C) / r

r = (667.8 x 10-3 Nm2/C) / (7.82 V)

r = 85.4 x 10-3 m

r = 85.4 mm

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