Question

m u= 0 + 9 Given a particle having charge q = +1.90 PC 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. Top view 11. What are the three external forces acting on the 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

Solutions:

11. The three external forces for charge field system are as follows:

(I) gravitational force (downward),

(ii) normal force (normal reaction force by table) (upward)

(iii) tension (tension of string) (towards point P)

Electric force is internal force for charge field system

12. Work done by each external force is zero here, As the displacent of the particle is always perpendicular to each external forces.

13. Electric potential difference = negative of work done by electric field per unit charge = (- work done)/(charge)

$Potential \ difference\ = \frac{-q*E*L(1-\cos\theta)}{q} \\ = -E*L(1-cos(75\degree)) =- 570.71 V$

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9 t 1.90HC - 1.10 m L 700v/m 9 = 750 = 100 my ELI-68) Work done and faksabiel lfarss -work done Charge = -EL [1-) 570. 7 V

14. Electric potential energy of the system is decreased and there is no any work done by external forces so, kinetic energy = - (potential difference * charge ) = 570.71*q = 570.71 V * 1.90 uC

Or,

$\frac{1}{2}mv^{2} = 570.71*1.90*10^{-6} \\ v^2 = \frac{2*570.71*1.90.10^{-6}}{m} = \frac{2*570.71*1.90.10^{-6}}{100*10^{-6}} = 21.68698 \\ \\ or,\ v = \sqrt{21.68698}\ = 4.657\ m/s$