Problem 14(6.25 points) A Hea ion is moving towards a proton (p) as shown in the

Question

Problem 14(6.25 points) A Hea ion is moving towards a proton (p) as shown in the figure. The position of the proton is fixed. The dashed cireles represent the equipotential surfaces around the proton and the electric potentials for each surface are shown in the figure. Initially, the Hea ion is so far away from the proton so that the electric potential at that point is essentially equal to zero. The initial kinetic energy of the He* ion is 20 eV. Which of the following correctly describes the motion of the Hea ion: (Note: charge for He2+ io: +3.2xtON C; charge of proton = + 1.6 x 1o» C?ev-1.6 ×10"9 J) 10V 20 V 40Y Het A) It will stop at point A and it will be bounced back It will stop at point B and it will be bounced back t will stop at point C and it will be bounced back Drit will not stop until it touches the proton Insufficient information to determine at which point the He ion will stop. Problem 15 (6.25 points) A metal sphere with a charge of +20 comes into contact with a metal sphere of identical size that has a charge of-40. The spheres are then separated, what are the charges on the spheres after they are separated? +2 0 th spheres have zero net charge Each sphere has a charge of -20 sphere has a charge of -Q sphere retains its original charge Problem 16 (6.25 points) A parallel plate capacitor can store an energy of 50 J when a voltage Vis applied across the two plates. Now, someone modifies the capacitor by reducing the distance (d inside the capacitor to half of its original value. A voltage 2Vis then used to charge up the capacitor. What will be the energy stored inside the capacitor after it is charged? the two plates ?more Cap (Hint: E = , CV2) 50J 50 sJ ) 200 J E) 400J

Explanation / Answer

14) A) it stops at point A and it will be bounced back.

Workdone = q*delta_V

change in kinetic energy = q*delta_V

==> delta_V = change in kE/q

= 20 eV/(2*e)

= 10 V

so, it stops at point A

15) C) Each sphere has a charge of -Q

charge each sphere = Qnet/2

= (+2*Q - 4*Q)/2

= -Q

16) E) 800 J

C = A*epsion/d

U = (1/2)*C*V^2

50 = (1/2)*C*V^2

C' = A*epsilon/(d/2)

= 2*A*epsilon/d

= 2*C

U' = (1/2)*C'*V'^2

= (1/2)*2*C*(2*V)^2

= 8*(1/2)*C*V^2

= 8*50

= 400 J

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