Question

Can someone answer c and d in detail, please??

Physics 8B 1- Charge and Coulomb's Law rev 4.0 . Two charges QA QB +Q are held fixed on the y-axis at (0, 3d) and (0,-3d). A third charge, Qc = +Q, is released from rest on the x-axis at (4d, 0). eA QB a) Q: Which way will Qc move? Explain b) Q: Describe the motion of Qc. Does it speed up? Slow down? Turn around? Where is it fastest? Etc. c) P: Find the force (mag. and dir.) on Qc- Q&P: Repeat a) -e) for the case where Q20.

Answer 1

Because Q_A, Q_B, Q_c all are +ve charges so force on Q_c due to Q_A and Q_B will be repulsive as shown in figure Now repulsive force F_A on Q_c due to Q_A is By coulombs's law - F_A = R Q_A Q_C/{squareroot (3d^2) + (4d)^2}^2 {because d_1 = squareroot (3d)^2 + (4d)^2 implies F_A = R Q times Q/(5d)^2 implies F_A = R Q^2/25 d^2 rightarrow (1) similarly force F_B on Q_C due to Q_B - F_B = R Q_B Q_C/d_2^2 = R times Q times Q/{squareroot (3d)^2 + (4d)^2 implies F_B = R Q^2/25 d^2 rightarrow (2) we can see from equation (1) and (2) that F_A = F_B = R Q^2/25 d^2 rightarrow (3) (only in magnitude) Now let us resolve F_A and F_B into horizontal and vertical components we see that vertical components F_B sin theta and F_A sin theta are equal and opposite so cancel each other, while horizontal components are in same direction