Particles 91 = - 75.8 MC, 92 = +90.6 MC, and 93 = -84.2 MC are in a line. Particles q1 and 92 are separated by 0.876 m and particles 92 and 93 are separated by 0.432 m. What is the net force on particle 93? Remember: Negative forces (-F) will point Left Positive forces (+F) will point Right

Answer:

The net force on particle q3 is -35.1 N (leftward)

Step-by-step explanation:

To calculate the net force on q3, we need to first calculate the individual forces between each pair of particles, using Coulomb's Law:

F = k * (q1*q2)/(r^2)

where F is the force, k is Coulomb's constant (9.0 x 10^9 N*m^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between them.

For q3, the forces due to q1 and q2 will be in opposite directions, so we need to subtract them to find the net force:

F_net = F(q2,q3) - F(q1,q3)

where F(q2,q3) is the force due to q2 on q3 and F(q1,q3) is the force due to q1 on q3.

Let's plug in the given values and calculate:

F(q1,q2) = k * (q1*q2)/(r^2) = 9.0 x 10^9 * (-75.8 x 10^-6 * 90.6 x 10^-6)/(0.876^2) = -7.08 N (leftward)

F(q2,q3) = k * (q2*q3)/(r^2) = 9.0 x 10^9 * (90.6 x 10^-6 * (-84.2 x 10^-6))/(0.432^2) = -42.2 N (leftward)

Therefore, the net force on q3 is:

F_net = F(q2,q3) - F(q1,q3) = -42.2 N - (-7.08 N) = -35.1 N (leftward)

So, the net force on particle q3 is -35.1 N (leftward).