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Please provide deep explaining of concepts, theories, and formulas that apply to each problem. I have a physics final coming up and I need to be prepared, these are the practice questions. thank you.

4. -6 points SerCP10 15.P.010. Calculate the magnitude and direction of the Coulomb force on each of the three charges shown in the figure below 6.00 uC charge magnitude 6.00 uC charge direction Select 1.50 uC charge magnitude 1.50 uC charge direcionSelect -2.00 C charge magnitude -2.00pC charge direction -Select- 1.50 :C -2.00m 8.00 ca 200 cm Need Help? 5. -2 points SerCP10 15.P011 MI.FB Three charges are arranged as shown in the figure below. Find the magnitude and direction of the electrostatic force on the charge q-4.78 nC at the origin. (Let r12 0.255 m. direction o counterclockwise from the +x-axis 6.00 nC 0,100 m -300 nC Need Help? asher 6. -12 A smal object of mass 3.60 g and charge-19.2 C is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field? direction 1-select- Need Help? SerCP10 15.P017 N/C

Explanation / Answer

4)Let q1 = 6 uC ; q2 = 1.5 uC and q3 = -2 uC

on q1 force due to q2 is:

F12 = k q1q2/r12^2

F12 = 9 x 10^9 x 6 x 10^-6 x 1.5 x 10^-6 / (0.03)^2 = 90 N

F13 = k q1q3/r13^2

F13 = 9 x 10^9 x 6 x 10^-6 x -2 x 10^-6 / (0.05)^2 = -43.2 N

Fnet1 = F12 + F13 = 90 - 43.2 = 46.8 N

Hence, Fnet1 = 46.8 N

on q2 force due to q1 is:

F12 = k q1q2/r12^2

F12 = 9 x 10^9 x 6 x 10^-6 x 1.5 x 10^-6 / (0.03)^2 = 90 N

F23 = k q2q3/r13^2

F23 = 9 x 10^9 x 1.5 x 10^-6 x -2 x 10^-6 / (0.05)^2 = -67.5 N

Fnet2 = F12 + F23 = 90 - 67.5 = 22.5 N

Hence, Fnet2 = 22.5 N

on q3 force due to q2 is:

F23 = k q2q3/r13^2

F23 = 9 x 10^9 x 1.5 x 10^-6 x -2 x 10^-6 / (0.05)^2 = -67.5 N

F13 = k q1q3/r13^2

F13 = 9 x 10^9 x 6 x 10^-6 x -2 x 10^-6 / (0.05)^2 = -43.2 N

Fnet1 = F12 + F13 = -67.5 - 43.2 = -110.7 N

Hence, Fnet3 = -110.7 N

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