Question

Calculate the speed of a proton after it accelerates from rest through a potential difference of 220 V . (answer in m/s) Calculate the speed of an electron after it accelerates from rest through a potential difference of 220 V . (answer in m/s) A point charge of -8.3 μC is at the origin. What is the electric potential at (3.0 m, 0)? What is the electric potential at ( - 3.0m,0)? What is the electric potential at (3.0m, - 3.0m)? (answer in V) How far must the point charges q1 = 7.10 μC and q2 = -23.0 μC be separated for the electric potential energy of the system to be -110 J ?   

Answer 1

1.

Equate, potential energy of electron with kinetic energy.

1/2*m*v² = q*V

v = sqrt(2*q*V/m)

v = sqrt(2*1.6*10^-19 *220/1.67*10^-27)

v = 2.05*10⁵ m/s

2.

Equate, potential energy of electron with kinetic energy.

1/2*m*v² = q*V

v = sqrt(2*q*V/m)

v = sqrt(2*1.6*10^-19 *220/9.1*10^-31)

v = 8.79*10⁶ m/s

3.

(a)

potential is scalar .

V = q/4*pi*o*d

V = -8.3*10^-6 /4*3.14*8.85*10^-12 * 3

V = -0.0248*10⁶

V = -24.8*10³V

(b)

potential is scalar so

potential at (3,0) = potential at (-3,0) = -24.8*10³ V

(c)

potential at (3,-3)

V = k*q/d

d = 32

V= 9*10⁹ *-8.3*10^-6/ 32

V = 17.60*10³ V