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. 01 0 (a) Find the electric field at the location of Q1, due to Q2 and Q3 E-<-2.16e5, - 2.88e50> x N/C (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. N/C (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 1027 kg for the mass of He2+.) m/s2

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Answer #1

Electric field due to a point charge Q at a distance r is given by

E=

In vector notation,

  vec{E} = rac{1}{4 pi epsilon _{0}} , rac{Q}{left | r ight |^{3}} , vec{r}

Where, vec{r} is vector starting from the charge and ending to the point (where electric field is to be calculated). and

4πε0

Given,  

  Q_{1} = 3 mu C

Q_{2} = 6 mu C

Q_{3} = -2 mu C

A(3,4) Q2

(a) The electric field due to charge Q_{2} and Q_{3} at the location of Q_{1} is shown below

A (3,4) Ев Ea 4 cm 3 cm

Here, E _{B} and E _{O} is electric field due to charge Q_{3} and Q_{2} respectively. And

  tan-1 36.86937

The magnitude of these electric fields can be calculated as

ー=337.5 x 105 N/C 4 x 10-2)2

The direction of this electric field is towards the positive y - direction. So,

Eo = (337.5 x 105 N/C) j

And

  1 Q3 2 × 10-6 04TAB2((4 x 10-2)2 +(3 x 10-2)2)

  72 x 105 N/C (only the magnitude because direction is already shown in the diagram)

In vector form vec{E} _{B} can be written as

vec{E} _{B} = -E_{B} , cos , heta , hat{j} + E_{B} , sin , heta , hat{i}

or,   vec{E} _{B} = -(72 imes 10^{5}) , cos , 37^{circ} , hat{j} + (72 imes 10^{5}) , sin , 37^{circ} , hat{i}

or, Ep--(57.6 x 105) J + (43.2 x 105) i

So, net electric field due to charge Q_{2} and Q_{3} at the location of Q_{1} is

  vec{E}_{net} = vec{E}_{O} + vec{E}_{B}

  (279.9 x 105) j+ (43.2x 105) ; (in N/m)

(b) So, the net force on Q_{1} due to these two charges is

  net 1 net

= 3 x 10-6 x ( (279.9 x 105) j + (43.2x 105) i

(83.97) j + (12.96) i

(c)  

EOA A-(3 ,4) EAG (01) AL 13 (93) AElactic vield at A dua to charge 3 Enc. → ElechicIjídd at A drue to charge. Q1 So→ , AB (3 xl02

OA 2 Nous So, net electric lidd at A OA net (423.6 x1o5)个t (60.3x(os)

(d)

Total charge on alpha particle = 2e = 2 x 1.602 x 10-19 C = 3.204 x 10-19 C

And total mass of alpha particle = 6.646 x 10-27 kg

So, total force on the alpha particle at A is

Ftotal = 3.204 x 10-19 x EA

3.204 x 10-1 x ((429.6x10(60.3 x 10) j

= (13.76 x 10-12) i+ (1.93 x 10-12) (in N)

The acceleration of the alpha particle is

r total mal pha a =

6.646 X 10-27 x ((13.76 x 101.93 x 10-12)j

(2.07 x 105(0.3 x 1015)j

For any doubt please comment and please give an up vote. Thank you.

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