Question

m Given a particle having charge q = +1.90 C and mass m = 100 mg is connected to a string that is L = 1.10 m long and tied to the pivot point P in the figure below. The particle, string, and pivot point all lie on a horizontal table. The particle is released from rest when the string makes an angle @= 75.0° with a uniform electric field of magnitude E = 700 V/m. u= 0 + 9 po el E 11. What are the three external forces acting on the Top view charge-field system 12. Find the work done by each of the EXTERNAL FORCES 13. What is the electric potential difference as the particle moves from the initial to the final position? 14. Determine the speed of the particle when the string is parallel to the electric field.

Answer 1

11)

Three external forces acting on the charge-field system are,

i) Gravitational force vertically downwards

Fg = mg

ii) Normal force $F_N$ , vertically upwards

iii) Electrical force $F_E$ , horizontally to right.

12)

Work done by gravitational force , because the gravitational force is perpendicular to the displacement.

W, = 0

Work done by normal force   , because the normal force is perpendicular to the displacement.

WN=0

Work done by electrical force

We = FE *(ry-ri) =qE * (L - L cosa)

We = 1.90 * 10-6* 700* (1.10 – 1.10 cos 75.0°) = 1.08 * 10-3 J

13)

Electric potential difference

WE AV = 700 1.10 - 1.10 cos 750 L - L cos @

14)

mv2 = 24V

29AV V = V m

Speed of particle is  

V = V 2 * 1.90 * 10-6 * 858.6 V = 5.71 m/s 100 * 10-6