A charge of #1 C# is at #(-5 ,-2 ,7 )# and a charge of #2 C# is at #(0,8,4) #. If both coordinates are in meters, what is the force between the charges?

1 Answer
Feb 13, 2016

Answer:

Force between charges is #1.343⋅10^8# Newton

Explanation:

Distance #d# between #(x_1,y_1,z_1)# and #(x_2,y_2,z_2)# is given by

#sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#.

In the instant case, this turns out to be #sqrt((0+5)^2+(8+2)^2+(4-7)^2)#, which simplifies to #sqrt(25+100+9)# or #sqrt134#.

The force between two charges #C_1# and #C_2# is given by #k*(C_1*C_2)/d^2#, where #k# is Coulomb's law constant for the medium between the charges and for air it is #9*10^9 Nm^2/C^2#, where charge is in Coulomb and distance in meters.

Assuming it to be so Force will be given by

#9*1*2/134*10^9# or #1.343*10^8# Newton.