A charge of +2C is at the origin. When a charge Q is placed at2m along the posit

Question

A charge of +2C is at the origin. When a charge Q is placed at2m along the positive axis, the electric field at 2m along thenegative x axis becomes zero. What is the value ofQ?        a.-3C     b.-6C       c.   - 7 C         d.   -8 C

8. Electrons in a particle beam each have a kinetic energy of3.2x10-17 J. What is the magnitude of the electric field that willstop these electrons in a distance of 0.1m?

a. 200 N/C       b.1000N/C     c. 2000 N/C    d. 4000N/C

9. A proton (+1.6x10^-19 C) moves 10cm along the direction of anelectric field os strength 3.0 N/C. The electrical potentialdifference between the proton’s initial and ending pointis:

a. 4.8x 10^-19 V      b.0.30V      c.0.033V     d. 30V

10. If an electron is accelerated from rest through a potentialdifference of 1200V, find its approximate velocity at the end ofthis process. ( e= 1.6x10^-19 C, m= 9.1x10^-31kg)

a. 1,0x10^7m/s           b.1.4x10^7m/s           c.2.1 x 10^7m/s         d. 2.5x10^7m/s

Explanation / Answer

Q at +2m then exactly balances/cancels out the electric field at-2m due to +2C at the origin. Thus, from E = Q/4or2 we have, E1 = [(2C)/{(40)(2*2)}](-i) = -E2 =-[(Q)/{(40)(2-(-2))*(2-(-2))}](-i) So, 2C/4 = -Q/16 So, Q = -8C Hence, (d) is the correct choice. ------------------------------------------------------------------------------ Speed of the electrons = v = (2*T/m) where T is their kinetic energy and m is the electronic mass. The deceleration caused by an electric field of magnitude E is: a = eE/m using Newton's second law. So, using the equation of motion under constant acceleration, v'2 - v2 = 2as So, |a| = |v2/2s| = eE/m So, E = mv2/2es = T/es =(3.2*10-17)/(1.602*10-19*0.1) N/C = 2000N/C Hence, option (c) is correct. ------------------------------------------------------------------------------ Electric potential difference = Electric field * displacement =3.0*0.10 V = 0.30 V Hence, option (b) is correct. ------------------------------------------------------------------------------ Change in kinetic energy and final kinetic energy are : T - 0 = T = e*V but T = mv2/2 So, v = (2eV/m) =[(2*1.6*10-19*1200)/(9.1*10-31)] m/s =2.054*107 m/s ˜ 2.1*107 m/s. Hence, option(c) is correct.

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