3. + 014 points | Previous Answers M14 13.5.048. My Notes + Ask Your Teacher Fie

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3. + 014 points | Previous Answers M14 13.5.048. My Notes + Ask Your Teacher Field and force with three charges At a particular moment, one negative and two positive charges are located as shown in the figure. Your answers to each part of this problem should be vectors. It helps a great deal to make a diagram with arrows representing the various electric field contributions, and then check the signs of your components against these arrows. Let Q1 = 1 pc, Q2 = 7 pc, and Q3 = -7 pc. 1 cm ---- (a) Find the electric field at the location of Qi, due to Q2 and Q3. E = (b) Use the electric field you calculated in part (a) to find the force on Q1. F = (c) find the electric field at location A, due to all three charges. N/C E = (d) An alpha particle (He2+, containing two protons and two neutrons) is released from rest at location A. Use your answer from part (c) to determine the initial acceleration of the alpha particle. (Use 6.646 x 10-27 kg for the mass of He2+.) a = m/s2

Explanation / Answer

We know that the electric field due to a point charge is given by:

E = kq/r^2

E2 = 9 x 10^9 x 7 x 10^-6/0.04^2 = 3.94 x 10^7 N/C

E3 = 9 x 10^9 x 7 x 10^-6/(0.04^2 + 0.03^2) = 2.52 x 10^7 N/C

theta = tan^-1(-4/3) = -53.1 deg = 306.9 deg

Ex = E2 cos90 + E3 cos306.9

Ex = 0 + 2.52 x 10^7 x cos306.9 = 1.51 x 10^7 N/C i

Ey = E2 sin90 + E3 sin306.9

Ey = 3.94 x 10^7 x 1 + 2.52 x 10^7 x sin306.9 = 3.94 x 10^7 - 2.02 x 10^7 = 1.92 x 10^7 N/C j

E = sqrt (Ex^2 + Ey^2) = sqrt (1.51^2 + 1.9^2) x 10^7 = 2.44 x 10^7 N/C

b)F = qE

F = 1 x 10^-6 x 2.33 x 10^7 = 23.3 N

Hence, F = 23.3 N

c)E1 = 9 x 10^9 x 1 x 10^-6/0.03^2 = 1 x 10^7 N/C

E2 = 9 x 10^9 x 7 x 10^-6/(0.04^2 + 0.03^2) = 2.52 x 10^7 N/C

theta1 = tan^-1(4/3) = 53.1 deg

E3 = 9 x 10^9 x 7 x 10^-6/(0.04^2) = 3.94 x 10^7 N/C

theta2 = 270 deg

Ex = 1 x 10^7 x cos0 + 2.52 x 10^7 x cos53.1 + 3.94 x 10^7 x cos270 = 2.51 x 10^7 N/C

Ey = 1 x 10^7 x sin0 + 2.52 x 10^7 sin53.1 + 3.94 x 10^7 x sin270 = -1.92 x 10^7 N/C

E = sqrt (Ex^2 + Ey^2)

E = sqrt (2.51^2 + -1.92^2) x 10^7 = 3.16 x 10^7 N/C

theta = tan^-1(-1.92/2.51) = -37.41 deg = 322.6 deg

d)F = ma => a = F/m

a = q E/m = 2 x 1.6 x 10^-19 x 3.16 x 10^7/(6.64 x 10^-27) = 1.52 x 10^15 m/s^2

Hence, a = 1.52 x 1015 m/s^2

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