Question

A point charge, Q = −0.3 µC and m = 3 × 10−16 kg, is moving through the field E = 30az V/m. Use Eq. (1) and Newton’s laws to develop the appropriate differential equations and solve them, subject to the initial conditions at t = 0, v = 3 × 105ax m/s at the origin. At t = 3 µs, find (a) the position P(x, y, z) of the charge; (b) the velocity v; (c) the kinetic energy of the charge.

Answer 1

a).
`p(x,y,z)=(0.90m,0,0.135m)`

b).
`v(x,y,z)=(3x10^5 a_(x),0,-9x10^4 a_(z))`

c).
`E_(k)=1.5x10^(-5)J`

Step-by-step explanation:

The Newton second law using to determinate position

a).

`F=m*a`

`F=m*a=m*(d^2Z)/(dt^2)`

`(dZ)/(dt)=v_(z)=(q*E)/(m)*t`

`z=(q*E)/(2*m)*t^2`

`z=\frac{(0.3x10{-6}*30}{2*(3x10^(-16))}`

`z=-1.5x10^(10)*t^2m`

So at t=3us

`z=-(1.5x10^(10))*(3x10^(-6))^2=-0.135m`

To x at t=3us

`x=v*t=3x10^5*(3x10^(-6))^2=0.90m`

`p(x,y,z)=(0.90m,0,0.135m)`

b).

Velocity now, derive the position can get the velocity of the function so

`z=(q*E)/(2*m)*t^2`

`(dz)/(dt)=(q*E)/(m)*t`

`v=(q*E)/(m)*t`

at t=3us

`v=(-0.3x10^(-6)*30)/(3x10^(-16))*3x10^(-6)=-9x10^4(m)/(s)`

so

`v=(3x10^5 a_(x),0,-9x10^4 a_(z))`

c).

Kinetic energy

`E_(k)=(1)/(2)*m*v^2`

`E_(k)=(1)/(2)*3x10^(-16)*(1.13x10^5)^2=1.5x10^(-5)J`