Motion of Charge Particle in Electric Field

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, $$W=Q\left( \overrightarrow{E}.\overrightarrow{r} \right)=Q\left( {{E}_{1}}{{r}_{1}}+{{E}_{2}}{{r}_{2}}+{{E}_{3}}{{r}_{3}} \right)$$.