Particles of charge +65, +48, and 95 C are placed in a line (Figure 1) . The cen
ID: 1406137 • Letter: P
Question
Particles of charge +65, +48, and 95 C are placed in a line (Figure 1) . The center one is L = 55 cm from each of the others.
Part A
Calculate the net force on the left charge due to the other two. Enter a positive value if the force is directed to the right and a negative value if the force is directed to the left.
F+65=
Part B
Calculate the net force on the center charge due to the other two. Enter a positive value if the force is directed to the right and a negative value if the force is directed to the left.
F+48=
Part C
Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right and a negative value if the force is directed to the left.
F-95=
Explanation / Answer
A)
F = K*q1 * (q2/r1^2 + q2/r2^2)
= (9*10^9)*(65*10^-6)* ((48*10^-6)/(0.55^2) + (-95*10^-6)/(2*0.55)^2)
=(0.585)* (48/(0.55^2) + (-95/(1.1)^2)
= 46.9 N
Here positive means force was repulsive. Hence force is directed towards left.
So you must enter :
Answer: -46.9 N
B)
F = K*q2 * (q1/r1^2 + q3/r2^2)
= (9*10^9)*(48*10^-6)* ((65*10^-6)/(0.55^2) + (-95*10^-6)/(0.55)^2)
=(0.432)(65*/(0.55^2) + (-95)/(0.55)^2)
= -42.8 N
Here positive means force was attractive. Hence force is directed towards right.
So you must enter :
Answer: 42.8 N
C)
F = K*q1 * (q2/r1^2 + q2/r2^2)
= (9*10^9)*(-95*10^-6)* ((48*10^-6)/(0.55^2) + 65*10^-6)/(2*0.55)^2)
=(-0.855)* (48/(0.55^2) + (65/(1.1)^2)
= -181.6 N
Here negative means force was attractive. Hence force is directed towards left.
So you must enter :
Answer: -181.6 N