Hi, Please give me the solutions and please explain to me how to do this problem

Question

Hi, Please give me the solutions and please explain to me how to do this problem as I have a feeling it will be on my next EXAM. THANKS!

Explanation / Answer

q1 = 1.5 µC, q2 = 3.0 µC, q3 = 4.5 µC, and q4 = 6.0 µC. (A) Determine the magnitude of the electric field E at the origin. (B) Determine the direction of the electric field E at the origin. Express your answer as an angle ?, measured counterclockwise from the +x direction. Now an electron of mass m = 9.11×10^-31 kg and charge q = -1.60 ×10^-19 C is placed at the origin and released from rest. (C) Calculate the magnitude the initial acceleration of the electron. (D) Determine the direction of the initial acceleration of the electron. (A) electric field at distance r is given by E = k*q/r^2 and it is a vector quantity whose direction, for positive charfe, is given by line joining position of charge to point of interest ( origin in this case). here we have electric field E1 = k*q1/(9*10^-4) by q1 towards left at origin. similarly, we have E2 = 2*E1 downwards, E3 = 3*E1 rightwards and E4 = 4*E1 upwards. adding them vectorically, we have .. E = 4.24264 * 10^6 N/C. (B) E1, E3 are on same x-axis and E2,E4 are on same Y-axis we have, E3>E1 and E4>E2. also, E3-E1 = E4-E2 so, direction of resultant is given by field acting towards +x and +y axis and inclined to them equally so, ? = 45* (C) q*E = m*a (by force equation) where, q = charge of electron, m = mass of electron, a = acceleration of electron so, a = q*E/m = -7.45958 * 10^18 m/s^2 (D) direction of initial acceleration of electron will be opposite to direction of field (as electron is negatively charged) so, its direction is towards south-west..at angle -45 with -x axis so, direction angle, ? = 225*

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