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(Solved): Two charges q and Q are fixed in place along the x-axis. Q = +3.8 nC is at x = -2.0 cm and unknown c ...


Two charges q and Q are fixed in place along the x-axis. Q = +3.8 nC is at x = -2.0 cm and unknown charge q are located at x = 2.0 cm.

a) It is found that the net electric field is zero at x = -0.76 . What is the magnitude and sign of charge q?

b) Sketch both the equipotential lines and the electric field for the system (you need to have at least 5 field lines for the smallest charge and 6 potential lines for each charge)

¢) A proton is fired at infinity towards g along the line of the two charges with speed 7.4x10^4 m/s. How close is the proton from charge q before it turns around?



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To solve this problem, we can use the principles of electrostatics and Coulomb's law. Coulomb's law states that the electric field due to a point charge is given by: E = k * (Q / r^2) where E is the electric field, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance from the charge. a) To find the magnitude and sign of charge q when the net electric field is zero at x = -0.76 cm, we can set up the equation for the net electric field: E_net = E_q + E_Q Since the net electric field is zero, we have: 0 = k * (q / (0.76 cm)^2) + k * (3.8 nC / (2.76 cm)^2) Simplifying the equation, we get: q / (0.76 cm)^2 = -3.8 nC / (2.76 cm)^2 Now we can solve for q: q = -3.8 nC * (0.76 cm)^2 / (2.76 cm)^2 Calculate the value to find the magnitude and sign of charge q.

Please refer to the solution in this step


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