Question

Two positive point charges 5.25 μC are fixed on the y-axis at y = 7.75 mm and y = -7.75 mm. A negative point charge -4.25 μC is located 10.0 mm on the x-axis.

What is the x-component of the force on the -4.25 μC charge?

What is the y-component of the force on the -4.25 μC charge?

What is the magnitude of the net force on the -4.25 μC charge if it is located at the origin?

Where must the -4.25 μC charge be place to obtain the maximum force on it from the 2 charges q?

What is that maximum force (magnitude only)?

Answer 1

Two +ve point charges 5.25 uC each

pos. of the charge (0,+7.75 mm) , (0, -7.75mm)

third charge = -4.25 uC at ( 10.0,0)

each +ve charge is at a distance r = sqrt(10² + 7.75² ) = 12.65 mm from the -ve charge

F = k q₁q₂ /r²  

The force on the -ve charge is directed along the line joining them

x-comp of the force F_x = F * cos(p)

= 9.0e+9 * 5.25e-6 *-4.25e-6 / 160.06e-6 * (10/12.65)

= -0.99 e-3 N from each charge

the x-comp. of force from each +ve charge is directed in the same direction and they add up

Total x-comp of the force = 2*-0.99e+3 = -1.98e+3 N

y-comp. of the force from each charge

F_y = F*Sin(p) = 9.0e+9 * 5.25e-6 *-4.25e-6 / 160.06e-6 * (7.75/12.65)

= -0.769e+3 N from each charge.

The y-comp of the forces from each +ve charge are directed in opposite directions and they cancel up

If the -ve charge is located at the origin. x-component of the force from each charge will be 0 Cos(90) =0

and the y-compos. are directed opposite and cancel up.

The net force on the -ve charge =0

The net force on the -ve charge

F_net = 2*k q₁q₂ /r² Cos(p)

r = y/sin(p)

F_net = 2*k q₁q₂ /y² * Sin²(p) *Cos(p)

the force depends only on the angle p

to maximise the force we differentiate wrt p and the dF/dxp =0

dF_net /dp = K d/dp ( Cos(p) - Cos³ (p) ) where K = 2*k q₁q₂ /y² is const.

=0

=> -Sin(p) +3Cos²(p) Sin(p) =0

Cos²(p) = 1/3

p = 54.73 dg , the force is maximum

F_net-max   = 2* 9.0e+9 * 5.25e-6 *-4.25e-6 / (7.75e-3)² * Cos(54.73)

= 3.86 e+3 N