Question

1)If an electron travels 0.350 m from an electron gun to a TV screen in 22.0 ns, what voltage was used to accelerate it? (Note that the voltage you obtain here is lower than actually used in TVs to avoid the necessity of relativistic corrections.)

2)A charge q₁ = 2.12

Answer 1

1) Apply, s = 0.5*a*t^2

a = 2*s/t^2

= 2*0.35/(22*10^-9)^2

= 1.446*10^15 m/s^2

Workdone = m*a*s

q*V = m*a*s

V = m*a*s/q

= 9.1*10^-31*1.446*10^15*0.35/1.6*10^-19

= 2879 volts

2)

A) Vnet = V1 + V2

= k*q1/d1 + k*q2/d2

= 9*10^9*2.12*10^-6/3*1.33/4) - 9*10^9*5.17*10^-6/(1.33/4)

= -1.21*10^5 volts

b)
here, |q1| < |q2|, so the point P where Enet is zero is close to q1 and it is left to Q1.

Let x is the distance from q1 to P.

F13 = F23

k*q1/(x)^2 = k*q2/(a+x)^2

q1/(x)^2 = q2/(a+x)^2

(a+x)/(x) = sqrt(q2/q1)

(a+x)/(x)= sqrt(5.17/2.12)

(a+x) = x*1.56

1.33+x = x*1.56

x = 1.33/0.56

= 2.375 m   from q1