Two point charges are located on the y -axis as follows: one charge q 1 = -1.70

Question

Two point charges are located on the y-axis as follows: one charge q1 = -1.70 nC located at y=-0.595 m , and a second charge q2 = 4.00 nC at the origin (y=0).

Part A

What is the magnitude of the total force exerted by these two charges on a third charge q3 = 5.50 nClocated at y3 = -0.430 m ?

Two point charges are located on the y-axis as follows: one charge q1 = -1.70 nC located at y=-0.595 m , and a second charge q2 = 4.00 nC at the origin (y=0).

Part A

What is the magnitude of the total force exerted by these two charges on a third charge q3 = 5.50 nClocated at y3 = -0.430 m ?

Explanation / Answer

One charge q1= -1.70nC located at y1= -0.595m ,

Second charge q2= 4nC at the origin (y=0) .

Third charge q3= 5.50nC placed at y3 = -0.430m

The force exerted by q1 will be attractive and force exerted by q2 will be repulsive .Thus both the forces will be in same direction ,both towards q1(both forces in negative y axis direction)

Distance between q3 and q1 = r1 = 0.595-0.430=0.165m
Distance between q3 and q2 = r2 = 0.430-0=0.430m

The permittivity of free space = eo = 8.85*10^(12) C^2/(N*m^2)

Total force = F = f1 + f2

f1=k*q1*q3/(r1)^2

f2=k*q2*q3/(r2)^2

so F=f1+f2

F=k*q3 [ (q1/ (r1)^2) + (q2/ (r2)^2) ]

F=(9*10^9)*(5.50*10^(-9))*[(1.70*10^(-9))/(.165)^2+(4*10^(-9))/(0.430)^2]

F=4.2*10^(-6) N

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