**Motion of
Charge Particle in Electric Field**

**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 its 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 in time
t, it reaches the point B where its 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\(\left( p \right)=mv\Rightarrow p=m\times \frac{QEt}{m}=QEt\),

\(p=m\times \sqrt{\frac{2Q\Delta V}{m}}=\sqrt{2mQ\Delta V}\),

**(iv) Kinetic Energy (K.E): **Kinetic energy gained by the particle in time
t is:

\(K.E=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\left( \frac{QEt}{m} \right)}^{2}}=\frac{{{Q}^{2}}{{E}^{2}}{{t}^{2}}}{2m}\),\(\Rightarrow K.E=\frac{1}{2}m\times \frac{2QV}{m}=Q\Delta V\),

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

Where, \(\Delta V=\) Potential difference between the two positions of charge Q

\(\Delta V=\overrightarrow{E}.\Delta \overrightarrow{r}=E\Delta r\cos \theta \) ; Where \(\theta \) = the angle between direction of electric field and direction of motion charge. If charge Q is given a displacement \(\overrightarrow{r}=\left( {{r}_{1}}\widehat{i}+{{r}_{2}}\widehat{j}+{{r}_{3}}\widehat{k} \right)\) in an electric field

\(\left( \overrightarrow{E} \right)=\left( {{E}_{1}}\widehat{i}+{{E}_{2}}\widehat{j}+{{E}_{3}}\widehat{k} \right)\). The work done is, \(W=Q\left( \overrightarrow{E}.\overrightarrow{r} \right)=Q\left( {{E}_{1}}{{r}_{1}}+{{E}_{2}}{{r}_{2}}+{{E}_{3}}{{r}_{3}} \right)\).