A particle (charge = +40 $\mu$C)is located on the x axis at the point x = -20 cm,and a second particle (charge = -50 $\mu$C)is placed on the x axis at x = +30 cm.What is the magnitude of the total electrostatic force on a third particle (charge = -4.0 $\mu$C)placed at the origin (x = 0)?
A)41 N
B)16 N
C)56 N
D)35 N
E)72 N

Question

Quizplus · Accepted Answer

We can use Coulomb's law to calculate the force between each pair of particles and then add them vectorially to find the total force on the third particle:

F1 = (k*q1*q3)/r1^2
F2 = (k*q2*q3)/r2^2

where k is the Coulomb constant, q1, q2, and q3 are the charges of the particles, and r1 and r2 are their distances from the origin.

Plugging in the values, we get:

F1 = (9*10^9 N*m^2/C^2)*(40*10^-6 C)*(-4*10^-6 C)/(0.2 m)^2 = -36 N
F2 = (9*10^9 N*m^2/C^2)*(50*10^-6 C)*(-4*10^-6 C)/(0.3 m)^2 = -20 N

Note that the negative sign indicates that the forces are attractive (since the charges have opposite signs).

To find the total force, we add these vectors:

Ftot = F1 + F2 = (-36 N) + (-20 N) = -56 N

Since the question asks for the magnitude of the force, we take the absolute value:

|Ftot| = 56 N

Therefore, the answer is C.

A Particle (Charge = +40 μ\muμ C)is Located on the X Axis at the Point X

A Particle (Charge = +40 $\mu$ C)is Located on the X Axis at the Point X