JEE Main & Advanced Physics Electrostatics & Capacitance Motion of Charge Particle in Electric Field

(1) When charged particle initially at rest is placed in the uniform field

Suppose a charge particle having charge Q and mass m is initially at rest in an electric field of strength E. The particle will experience an electric force which causes it's motion.

(i) Force and acceleration : The force experienced by the charged particle is \[F=QE\].

Acceleration produced by this force is \[a=\frac{F}{m}=\frac{QE}{m}\]

(ii) Velocity : Suppose at point A particle is at rest and in time t, it reaches the point B  where it's velocity becomes v. Also if \[\Delta V=\] Potential difference between A and B, S = Separation between A and B

\[\Rightarrow \]\[v=\frac{QEt}{m}\]\[=\sqrt{\frac{2Q\Delta V}{m}}\]

(iii) Momentum : Momentum\[p=mv,\,\,p=m\times \frac{QEt}{m}=QEt\]

or  \[p=m\times \sqrt{\frac{2Q\Delta V}{m}}=\sqrt{2mQ\Delta V}\]

(iv) Kinetic energy : Kinetic energy gained by the particle in time t is \[K=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\left( \frac{QEt}{m} \right)}^{2}}=\frac{{{Q}^{2}}{{E}^{2}}{{t}^{2}}}{2m}\]

or \[K=\frac{1}{2}m\times \frac{2QV}{m}=Q\Delta V\]

(v) Work done : According to work energy theorem we can say that gain in kinetic energy = work done in displacement of charge i.e. \[W=Q\Delta V\]

where \[\Delta V=\] Potential difference between the two position of charge Q. (\[\Delta V=\overrightarrow{E\,}.\Delta \overrightarrow{\,r\,}=E\Delta r\,\cos \theta \] where \[\theta \] is the angle between direction of electric field and direction of motion of charge).

If charge Q is given a displacement \[\overrightarrow{\,r\,}=({{r}_{1}}\hat{i}+{{r}_{2}}\hat{j}+{{r}_{3}}\hat{k})\] in an electric field \[\overrightarrow{\,E\,}=({{E}_{1}}\hat{i}+{{E}_{2}}\hat{j}+{{E}_{3}}\hat{k}).\] the work done is \[W=Q(\overrightarrow{\,E\,}.\overrightarrow{\,r\,})=Q({{E}_{1}}{{r}_{1}}+{{E}_{2}}{{r}_{2}}+{{E}_{3}}{{r}_{3}})\].

Work done in displacing a charge in an electric field is path independent.

(2) When a charged particle enters with an initial velocity at right angle to the uniform field

When charged particle enters perpendicularly in an electric field, it describe a parabolic path as shown

(i) Equation of trajectory : Throughout the motion particle has uniform velocity along x-axis and horizontal displacement (x) is given by the equation \[x=ut\]

Since the motion of the particle is accelerated along y-axis

So \[y=\frac{1}{2}\left( \frac{QE}{m} \right)\,{{\left( \frac{x}{u} \right)}^{2}}\]; this is the equation of parabola which shows \[y\propto {{x}^{2}}\]

(ii) Velocity at any instant : At any instant t, \[{{v}_{x}}=u\] and \[{{v}_{y}}=\frac{QEt}{m}\] so \[v=\,|\overrightarrow{\,v}|\,=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{{{u}^{2}}+\frac{{{Q}^{2}}{{E}^{2}}{{t}^{2}}}{{{m}^{2}}}}\]

If \[\beta \] is the angle made by v with x-axis than \[\tan \beta =\frac{{{v}_{y}}}{{{v}_{x}}}=\frac{QEt}{mu}\].