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A negative charge of -0.0005 C exerts an attractive force of 9.0 N on a second charge that is 10 m away. What is the magnitude of the second charge?

q1= -0.0005 C

q2= ?

F=9.0 N

D= 10m

 

K=9.0 x 10^9 N x m^2/C^2

 

F=Kq1q2

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        D^2

 May 6, 2021
 #1
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Coulomb's Law

 

F  = kc  q1 q2 / (r^2)

 

9 = (9 x109)  (.0005) (q2) / ( 102)       N

 

 

q2 =  ( 9* 102)  / [(9x109)(.0005)]   C       Since it is an ATTRACTIVE force q2 will be positive (since q1 is negative)     You can finish ! cheeky

 May 6, 2021

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