Field and force with three charges At a particular moment, one negative and two

Question

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 = 2 µC, Q2 = 7 µC, and Q3 = -7 µC.

(a) Find the electric field at the location of Q1, due to Q2 and Q3.

(b) Use the electric field you calculated in part (a) to find the force on Q1.

(c) Find the electric field at location A, due to all three charges.

(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 1027 kg for the mass of He2+.)

Explanation / Answer

The way to do these is:
1 Find the angles of the fields at the point for the various charges:
at Q1: q2's field points at 90, q3's points at arctan(4/3) = 53.1° beneath the horizontal = 306.9°
This is because + charge fields point radially outward in all directions, - charge fields point radially inward in all directions.

2 Find distance squared and plug into E = kq/r²
E2 =k*6e-6/0.04² = 33,703,319 N/C at 90°
E3 = k*5e-6/(0.03²+0.04²) = 17,975,104 N/C at 306.9°

3. Sum vector fields at the point (q1 in this case)
Sum horizontal components
E2*cos90 + E3*cos306.9 = 10,785,062 N/C

Sum vertical components
E2*sin90 + E3*sin306.9 = 19,323,236 N/C

E1 = 22,129,280 N/C at 60.83° <-------------------- (a)

F = q*E = 2e-6*22,129,280 = 44.2N at 60.83° <--------------- (b)

(c)
At point a
E1 = k*3e-6/0.03² = at 29,958,506 N/C at 0°
E2 = k*6e-6/(0.03²+0.04²) = 21,570,124 N/C at arctan(4/3) = 53.1°
E3 =k*5e-6/0.04² = 28,086,099 N/C at 270°
Sum horizontal components
E1*cos0 + E2*cos53.1 + E3*cos270 = 42,900,581 N/C
Sum vertical components
E1*sin0 + E2*sin53.1 + E3*sin270 = -10,830,000 N/C
Ea = 44,246,454 at 345.8° <-------------------- (c)
(d)
F = m*a => a = F/m = 44,246,454*3.2e-19/6.646e-27 = 2.1304e15m/s² <---- (d)

there may be some calculation error,but concept is right

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