Two charges of 7 C 7C and 5 C5C are positioned on a line at points 2 2 and -4 −4, respectively. What is the net force on a charge of -4 C−4C at 0 0?

1 Answer
P dilip_k
Mar 6, 2016

=51.75*10^9N=51.75⋅109N

Explanation:

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By Coulomb's law we Know that force between two charges Q and q situated at a distance r is given by the formula F=k(qQ)/r^2F=kqQr2,where k =Coulomb's constant=9*10^9Nm^2C^-29⋅109Nm2C−2

Applying this formula the attractive force of 7C on -4C
=9*10^9*4*7/2^2N=63*10^9N9⋅109⋅4⋅722N=63⋅109N
The attractive force of 5C on -4C =9*10^9*4*5/4^2N=11.25*10^9N9⋅109⋅4⋅542N=11.25⋅109N

These two forces are oppositely directed so the net force will be towards right and its magnitude =(63-11.25)*10^9N=51.75*10^9N=(63−11.25)⋅109N=51.75⋅109N

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