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

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Physics

If the radius of the equipotential surface of point charge is15.2 m at a potential of 2.30 kV, what is the magnitude of thepoint charge creating the potential?

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4.3 Ratings (711 Votes)

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!          


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