Question

A charge of -7.00 μC is located on the y axis at y = 3.00 cm. A charge of 8.00 μC is located on the y axis at y = -3.00 cm. Find the force on a -1.47 μC charge on the x axis at x = 8.50 cm. Give only the magnitude of the force.

Answer 1

direction of force on -1.47 uC carge due to -7 uC charge is repulsive in nature.

direction of the vector=(0.085,0)-(0,0.03)=(0.085,-0.03)

distance =sqrt(0.085^2+0.03^2)=0.090139 m

then unit vector along this direction=(0.085,-0.03)/0.090139=(0.943,-0.3328)

force magnitude=k*q1*q2/r^2

where k=9*10^9

q1=7 uC

q2=1.47 uC

r=0.090139 m

force magnitude=9*10^9*7*10^(-6)*1.47*10^(-6)/0.090139^2=11.398 N

in vector form, force=11.398*(0.943,-0.3328) N

force due to 8 uC on -1.47 uC:

force is attractive in nature.

direction of force=(0,-0.03)-(0.085,0)=(-0.085,-0.03)

distance=sqrt(0.085^2+0.03^2)=0.090139 m

then unit vector along this direction=(-0.085,-0.03)/0.090139=(-0.943,-0.3328)

force magnitude=k*q1*q2/r^2

=9*10^9**10^(-6)*1.47*10^(-6)/0.090139^2=13.026 N

force in vector form=13.026*(-0.943,-0.3328) N

total force=11.398*(0.943,-0.3328)+13.026*(-0.943,-0.3328)=(-1.5352,-8.1283) N

magnitude of force=sqrt(1.5352^2+8.1283^2)=8.272 N