If the radius of the equipotential surface of point charge is 15.2 m at a potential...

If the radius of the equipotential surface of point charge is 15.2 m at a potential of 2.30 kV, what is the magnitude of the point charge creating the potential?

Solutions

Expert Solution

Note that

V = kq / r

Which implies that

q = V r / k

where

k =   8.99E+09   N m^2/C^2
V = the electric potential =    2300   N/C
r = the distance from the source charge   15.2   m


Thus,

q =    3.89E-06   C   [ANSWER]

DONE! It was nice working with you!

genius_generous answered 1 year ago

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