Question

Three identical conducting spheres are configured as shown. The charges on each sphere are q1=e, q2=-3e and q3=-2e where e = 1.6x10^-19 C. The distance d is 1.0 nm and q3 is a distnace 2d from the origin.

1. Three identical conducting spheres are configured as shown to the right. The charges on each sphere are q1 e, q2-3e and q3-2e where e 1.60 x 10-19 C. The distance d is 1.00 nm and q3 is a distance of 2d from the origin. 91 (a) What is the force, F3, on q3 from the other two charges? 43 (b) What is the electric field, E, at the origin? (c) An electron is placed at the origin. What will the is released? (d) Where will the electron travel, the top right, top left, bottom right or bottom left? q2 Maeiuede f is aceleration be immeadiatly after i (e) A dipole with dipole moment p = 1.00 x 10-10 C.m e is placed at the origin. How much work is required to turn it 180 degrees around? (f) The spheres q1 and q2 are briefly brought into contact and then returned to their original positions. What is E at the origin now?

Answer 1

1.The force due to charge q₁ on charge q₃ =$F_{1}=q_{1}q_{3}/4\pi \epsilon _{0}r_{1}^{2}=2*(1.6*10^{-19})^{2}*9*10^{9}/(d^{2}+4d^{2})=-1.8*10^{-11} N.$

The force due to charge q₂ on charge q₃=$F_{2}=q_{2}q_{3}/4\pi \epsilon _{0}r_{2}^{2}=6*(1.6*10^{-19})^{2}*9*10^{9}/(d^{2}+4d^{2})=5.5*10^{-11} N.$

The x component of the force F₁ due to charge q₁=$F_{1}cos\theta =1.8*10^{-11}*2d/\sqrt{5d^{2}}=1.6*10^{-11}N.$

The y component of force F₁=$F_{1}*d/\sqrt{5d^{2}}=0.8*10^{-11}N.$

The x component of the force F₂ due to charge q₂=$F_{2}cos\theta =5.5*10^{-11}*2d/\sqrt{5d^{2}}=4.9*10^{-11} N.$

The y component of the force F₂ =$F_{1}*d/\sqrt{5d^{2}} =2.5*10^{-11} N.$

The x component of the net force=(4.9-1.6)*10^-11 N=3.3*10^-11 N.

The y component of the net force=(0.8+2.5)*10^-11 N=3.3*10^-11 N.

The magnitude of the net force $F_{3}=3.3\sqrt{2}*10^{-11} N=4.7*10^{-11}N$ making an angle tan^-1(3.3*10^-11/3.3*10^-11)=45⁰ anticlockwise with the positive x-axis.

b)Place a unit positive charge at the origin.

The force F₁ on this charge due to charge q₁=$q_{1}*1/4\pi \epsilon _{0}d^{2}=14.4*10^{8} N$ pointing upwards.

The force F₂ on this charge due to charge q₂=$q_{2}*1/4\pi \epsilon _{0}d^{2}=-43.2*10^{8} N$ pointing downwards.

The total force due to charges q₁ and q₂ on the unit positive charge=(43.2-14.4)*10⁸ N=28.8*10⁸ N pointing downwards.

The force F₃ due to charge q₃ on the unit positive charge=$q_{3}*1/4\pi \epsilon _{0}(2d)^{2}=7.2*10^{8} N$ pointing to the right.

Therefore the electric field at the origin $\vec{E}=\sqrt{28.8^{2}+7.2^{2}}*10^{8}=29.7*10^{8} N$ at an angle tan^-1(28.8/7.2)=tan^-1(4)=76⁰ clockwise with the positive x-axis.

c) The magnitude of the electron's acceleration=F/m=29.7*10⁸/9.11*10^-31=3.3*10³⁹ m/s².

d) The electron will travel to the top left.

e)The work required $W=2pE_{x}=2*7.2*10^{8}*10^{10}=0.14 J$ where E_x is the component of the electric field at the origin.

f)Now the charges on the spheres are both the same and equal to (-3e+e)/2=-e.Then the net force due to both these charges on a unit positive charge at the origin=0 N.

Thus the electric field at the origin =7.2*10⁸ N/C pointing to the right.