Question 8.1: A charged particle of mass 2 kg and charge 3 C starts at poi...

(a) This is an initial-value problem because initial values are given. According to Newton’s second law of motion

F=ma=QE

where a is the acceleration of the particle. Hence

a=\frac{QE}{m}=\frac{3}{2}(12a_{x}+10a_{y})=18a_{x}+15a_{y} m/s^{2}

a=\frac{du}{dt}=\frac{d}{dt}(u_{x},u_{y},u_{z})=18a_{x}+15a_{y}

(b) Equating components and then integrating, we obtain

\frac{du_{x}}{dt}=18\rightarrow u_{x}=18t+A                              (8.1.1)

\frac{du_{y}}{dt}=15\rightarrow u_{y}=15t+B                             (8.1.2)

\frac{du_{z}}{dt}=0\rightarrow u_{z}=C                                            (8.1.3)

where A, B, and C are integration constants. But at t=0,  u=4a_{x}+3a_{z}. Hence,

u_{x}(t=0)=4\rightarrow 4=0+A   or   A=4

u_{y}(t=0)=0\rightarrow 0=0+B   or   B=0

u_{z}(t=0)=3\rightarrow 3=C

Substituting the values of A, B, and C into eqs. (8.1.1) to (8.1.3) gives

u(t)=(u_{x},u_{y},u_{z})=(18t+4,15t,3)

Hence

u(t=1s)=22a_{x}+15a_{y}+3a_{z}m/s

(c) Kinetic energy

(K.E.)=\frac{1}{2}m|u|^{2}=\frac{1}{2}(2)(22^{2}+15^{2}+3^{2})=718J

(d)

u=\frac{dl}{dt}=\frac{d}{dt}(x,y,z)=(18t+4,15t,3)

Equating components yields

\frac{dx}{dt}=u_{x}=18t+4\rightarrow x=9t^{2}+4t+A_{1}                             (8.1.4)

\frac{dy}{dt}=u_{y}=15t \rightarrow y=7.5t^{2}+B_{1}                                       (8.1.5)

\frac{dz}{dt}=u_{z}=3\rightarrow z=3t+C_{1}                                                    (8.1.6)

At t=0,  (x,y,z)=(1,-2,0); hence

x(t=0)=1\rightarrow 1=0+A_{1}   or   A_{1}=1

y(t=0)=-2\rightarrow -2=0+B_{1}   or   B_{1}=-2

z(t=0)=0\rightarrow 0=0+C_{1}   or   C_{1}=0

Substituting the values of A_{1},  B_{1}, and C_{1} into eqs. (8.1.4) to (8.1.6), we obtain

(x,y,z)=(9t^{2}+4t+1,7.5t^{2}-2,3t)                            (8.1.7)

Hence, at t=1,  (x,y,z)=(14,5.5,3).

By eliminating t in eq. (8.1.7), the motion of the particle may be described in terms of x, y, and z.