A charge of #-2 C# is at #(6 , -5 )# and a charge of #-3 C# is at #(3, 5 ) #. If both coordinates are in meters, what is the force between the charges?

1 Answer
Dec 30, 2015

Answer:

#F=4.95*10^8N#

Explanation:

The distance formula for Cartesian coordinates is

#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2#
Where #x_1, y_1#, and#x_2, y_2# are the Cartesian coordinates of two points respectively.
Let #(x_1,y_1)# represent #(6,-5)# and #(x_2,y_2)# represent #(3,5)#.

#implies d=sqrt((3-6)^2+(5-(-5))^2)#
#implies d=sqrt((-3)^2+(10)^2#
#implies d=sqrt(9+100)#
#implies d=sqrt(109)#

Hence the distance between the two charges is #sqrt(109)#.

The electrostatic force between two charges is given by
#F=(kq_1q_2)/r^2#
Where #F# is the force between the charges, #k# is the constant

and its value is #9*10^9 Nm^2/C^2#,#q_1# and #q_2# are the

magnitudes of the charges and #r# is the distance between the two charges.

Here #F=??#, #k=9*10^9Nm^2/C^2#, #q_1=-2C#, #q_2=-3C# and #r=d=sqrt109#.

#implies F=(9*10^9(-2)(-3))/(sqrt109)^2=(9*10^9*6)/109=(54*10^9)/109=0.495*10^9N=4.95*10^8N#
#implies F=4.95*10^8N#

Since the force is positive therefore the force is repullsive.