A constant electric field of E = -15000 N/C exists betweeen two large oppositely

Question

A constant electric field of E = -15000 N/C exists betweeen two large oppositely charged plates placed small distance d = 10 mm apart. a. The left plate has a potential of V_1 = 0 V. Determine the potential of the right plate, V_2. b. Points A and B are 2 mm as shown in the figure. What is V_A - V-B? c. Calculate -v_A - V_C. d. A particle at point C has velocity upsilon = -6 x 10^6 m/s i and slows down as it moves toward point A. Is the particle a proton or an electron? e. What is the speed of the particle at point A? f. How far from the left plate is the particle when it turns around and starts moving in the +i direction?

Explanation / Answer

1. direction of field is to the right and field is constant hence

deltaV = E.d

and d is the distance vector between two point points in the x directon.

hence V2 - V1 = (15000 V/m)(10 x 10^-3 m) = 150 Volt

b) distance between A and B along x axis =0

hence VA- VB = 0

c) VA - VC = (15000)(8 - 3) x 10^-3 = 75 volt

d) velocity is in -ve x direction and it is slowing doen that means

force is in +ve x direction .

and F = q E

E and F vector , both are in +ve x direction

hence q is positive.
Ans. Proton

e. change in KE = q deltaV

m vf^2 /2 - m vi^2 / 2 = (1.6 x 10^-19) (-75)

(9.109 x 10^-31) ( vf^2 - ( 6 x 10^6)^2) / 2 = 120 x 10^-19

( vf^2 - ( 6 x 10^6)^2) = 2.635 x 10^13

vf = 3.11 x 10^6 m/s

f. particle will turn around just after vf = 0

0 - (9.109 x 10^-31)(6 x 10^6)^2 / 2 = (1.6 x 10^-19) deltaV

deltaV = - 102.48 Volt = E.d

d = 102.48 / 15000 = 6.83 mm distace from point C

distance from left plate = 8 - 6.83 = 1.17 mm

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