What is the magnitude of the force F on the -10 nC charge in the figure(Figure 1

Question

What is the magnitude of the force F  on the -10 nC charge in the figure(Figure 1) ?

Part A What is the magnitude of the force F on the -10 nC charge in the figure(Figure 1)? Express your answer using two significant figures. What is the direction of the force F on the -10 nC charge in the figure? Give your answer as an angle measured cw or ccw (specify which) from the +x-axis. Express your answer using two significant figures. Enter positive value f the angle is counterclockwise and negative value if the angle is clockwise.

Explanation / Answer

Force on -10nC due to -5.0nC = k * 5*10 n^2 / (1 *10^-2)^2
F1 = 8.9 * 10^9 * 5*10*(10^-9)^2 / (1*10^-2)^2
F1 = 4.45 * 10^-3 N (towards -ve Y axis)

Distance r between -10n and 15n charge = sqrt(3^2 + 1^2) = 3.16 cm = 3.16*10^-2 m

Force on -10nC due to 15nC = k * 15*10 *n2 / (3.16 *10^-2)^2
F2 = 8.9 * 10^9 *15* 10 * (10^-9)^2 / (3.16 *10^-2)^2
F2 = 1.34 * 10^-3 N

sin () = 1/3.16
=sin^-1(1/3.16)  
=18.45o
F2y = F2 sin() (towards + ve Yaxis)
F2y = 1.34 * 10^-3 * sin(18.45o)
F2y = 4.241 * 10^-4 N

F2x = F2 cos() (towards -ve Xaxis)
F2x = 1.34 * 10^-3 * cos(18.45o)
F2x = 1.27 * 10^-3 N

Fy = F1 - F2y
Fy = 4.45 * 10^-3 N - 4.241 * 10^-4 N
Fy = 4.026 * 10^-3 N (towards -ve Yaxis)

Fx = 1.27 * 10^-3 N (towards -ve Xaxis)

F = sqrt(Fx^2 + Fy^2)
F = sqrt((1.27*10^-3)^2 + (4.026*10^-3)^2)
F = 4.22 * 10^-3 N

Magnitude of Force F = 4.2* 10^-3 N

tan() = 1.27 * 10^-3 /4.026 * 10^-3
= 17.51o

Direction is = 107.51 clockwise from +ve Xaxis

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