Motion of Charged Particle in Uniform Electric Field – Theory, Formula and Examples

17.6 MOTION OF CHARGE IN UNIFORM ELECTRIC FIELD

17.6.1 Charged Particles Moving in Direction Parallel to Uniform Electric Field

1. The uniform electric field

A uniform electric field is produced by a pair of flat parallel metal plates separated by a distance d. One plate is applied a positive voltage +V while the other is earthed, as shown in Figure 17.46. We have

(a) Field direction

The electric field lines are parallel, evenly spaced and point away from the positively charged plate.

(b) Magnitude of electric field strength

The magnitude of the electric field strength Eis given by

$E=\frac{V}{d}$

2. Direction of electric force

The electric force F_eacting on a particle in the field

(a) always points parallel to the field lines and towards the positively charged plate if the particle carries a negative charge

(b) always points parallel to the field lines and away from the positively charged plate if the particle carries a positive charge

 3. Acceleration of particle

Suppose a particle of mass mand carrying a charge qis able to move freely inside a uniform electric field of strength E. Then it will be acted on by a constant electric force $F_e$.

The magnitude of the electric force is given by

$F_e=ma$

But $F_e$ is also given by

$F_e=qE$

Hence we have

$ma=qE$ 

$a=\frac{q}{m}E$

$a=\frac{q}{m} \frac{V}{d}$

The direction of the acceleration vector apoints in the same direction as $F_e$.

4. Equations of motion

The charged particle moves along a linear path in a direction which is parallel to the electric force direction. We can apply the equations for a particle moving with constant acceleration

$v=u+at$

$s=ut+\frac{1}{2}at^2$

$v^2=u^2+2as$

EXAMPLE 17.26

A p.d. of 100 V is applied across a pair of flat parallel metal plates Aand B, separated by a distance of 1.0 cm. An electron emerges from plate Aat speed $5.0×10^6 \ m \ s^{-1}$ and moves towards Bin a linear path which is perpendicular to B. Determine

(a) the acceleration of the electron
(b) the speed of the electron when it reaches B
(c) the time taken by the electron to travel to B

For electron, $e/m_e =-1.76 \times 10^{11} \ C \ kg^{-1}$

Answer

(a) $E=\frac{V}{d}$

$E=\frac{100}{0.01}$

$E=1.0 \times 10^4 \ V \ m^{-1}$

Acceleration

$a= \frac{qE}{m}$

$a=\frac{1.76 \times 10^{11}}{1.0 \times 10^4}$

$a=1.76 \times 10^{15} \ m \ s^{-2}$

(b) $v^2=u^2+2as$

$=(5.0 \times 10^6 )^2+2(1.76 \times 10^{15})(0.01)$

$v=7.8 \times 10^6 \ m \ s^{-1}$

(c) $v=u+at$

$7.8 \times 10^6=5.0 \times 10^6+(1.76 \times 10^{15})t$

$t=1.6 \times 10^{-9} \ s$

17.6.2 Charged Particle Entering Uniform Electric Field at 90° to the Field

1. The electric field

The uniform electric field is produced by applying a p.d. Vacross a pair of flat parallel metal plates $P_1$ and $P_2$, which are separated by a distance d, as shown in Figure 17.47(a). Edge effects are neglected. The magnitude of the field strength Eis given by

$E=\frac{V}{d}$

2. Particle motion in field

A negatively charged particle travelling in a horizontal direction at constant speed uenters the electric field. The angle between the velocity and the field is 90°. We can determine the following for the particle moving inside the field.

(a) Force acting on particle

Once the negatively charged particle enters the electric field, it will immediately experience an electric force acting on it. This force has

(i) a direction which always points perpendicularly towards the positively charged metal plate $P_1$

(ii) a magnitude $F_e$ given by

$F_e=qE =q\frac{V}{d}$

where $q$ is the charge carried by the particle.

(b) Acceleration of particle

When the particle moves inside the field, the velocity is going to change continuously because it is acted on by an electric force. The magnitude increases with time and the direction of motion also changes. Hence, the particle has an acceleration $a$, given by

$a=\frac{F_e}{m}$

where mis the mass of the particle. The direction of aalways points perpendicularly towards the positively charged plate $P_1$.

(c) Path of motion

We can show that the particle follows a parabolic path in the field.

(d) Instantaneous velocity and its components

The instantaneous velocity $v$ of the charged particle in the field is no longer horizontal because the particle is following a parabolic path. Hence, at any instant it is going to have a  horizontal velocity component $v_x$ and a vertical velocity component $v_y$

(i) No external force acts in the horizontal direction (neglect friction), therefore

$v_x=u$

where $u$ is the initial speed of the particle before is enters the field.

(ii) Considering the vertical motion

$v_y=u_y+at$

But $u_y=0$, and so we have         $v_y=at$

(iii) The instantaneous velocity $v$ is given by 

$v=\sqrt{v_x^2+v_y^2}$

The vector diagram for $v$ is shown in Figure 17.47(a).

(e) Direction of instantaneous velocity

Let the angle between $v$ and the horizontal be θ. Then we have

$tan⁡ \ \theta=\frac{v_y}{v_x}=\frac{at}{u}$ 

(f) Time taken to cross the field completely

The time T taken by the particle to completely travel through the field is given by

 $T=\frac{L}{u}$

where L is the length of each of the metal plates.

EXAMPLE 17.27

Figure 17.48 The Y-plates found inside a cathode ray oscilloscope (CRO)

Figure 17.48 shows the Y-plates found inside a cathode ray oscilloscope (CRO). An electron moving at a constant speed of $1.5 \times 10^{7} \ m \ s^{-1}$ enters the vertical uniform electric field existing between the plates at an angle of 90°. A p.d. is applied across the plates, with the top plate charged positively at +80 V relative to the bottom plate.

(a) Copy Figure 17.48 and draw the path followed by the electron in between the plates and outside the plates.
(b) Determine the magnitude and direction of the acceleration of the electron in between the plates.
(c) Determine the magnitude and direction of the velocity of the electron when it has passed through the space between the deflecting plates.

$(e/m_e =-1.76 \times 10^{11} \ C \ kg^{-1} )$

Answer

(a) The path of the electron is shown in Figure 17.49.

Figure 17.49 The path of an electron in the Y-plates inside a cathode ray oscilloscope (CRO) 

(b) Refer to Figure 17.50

Figure 17.50

$F_e=q\frac{V}{d}$

$F_e=ma$

$a=\left(\frac{q}{m}\right) \left(\frac{V}{d}\right)$

$a=(1.76 \times 10^{11}) \left(\frac{80}{20 \times 10^{-3}}\right)$

$a=7.0 \times 10^{14} \ m \ s^{-2}$

(c) Refer to Figure 17.51.

The magnitude of velocity is given by

$v=\sqrt{v_x^2+v_y^2}$

$v_x=1.5 \times 10^7 \ m \ s^{-1}$

$v_y=at$

$t=\frac{L}{v_x}$ 

$t=\frac{60 \times 10^{-3}}{1.5 \times 10^7}=4.0 \times 10^{-9} \ s$

$v_y=(7.0 \times 10^{14})(4.0 \times 10^{-9})$

$v_y=0.28 \times 10^7 \ m \ s^{-1}$

$v=10^7 \sqrt{1.5^2+0.28^2}$

$v=1.53 \times 10^7 \ m \ s^{-1}$

$v =1.5 \times 10^7 \ m \ s^{-1}$

Direction

$tan \ \theta =\frac{v_y}{v_x}$ 

$tan⁡ \ \theta=\frac{0.28}{1.5}$

$\theta =10.6^0$  to the horizontal

17.6.3 Charged Particle Entering Uniform Electric Field at Less than 90°

1. The electric field

The uniform electric field is produced by applying a p.d. V across a pair of flat parallel metal plates which are separated by a distance d, as shown in Figure 17.53(a). Edge effects are neglected. The magnitude of the field strength E is given by

$E=\frac{V}{d}$

2. Particle motion in field

A positively charged particle travelling at constant speed u enters the electric field through a slit A in the bottom plate at angle θ to the plate.

We have the following:

(a) Electric force

When the positively charged particle enters the field, it will experience a repulsive electric force F_ebecause it moves towards the top plate which is also positively charged. The magnitude of the force is

$F_e=q\left(\frac{V}{d}\right)}$

The direction of the force is always vertically downwards.

(b) Path of motion

The particle follows a parabolic path inside the electric field.

(c) Instantaneous velocity in the field

Suppose that the particle enters the field through slit A and after time t it reaches point C. The velocity at that instant is v, which is inclined to the horizontal at angle φ. It has a horizontal component and a vertical component, as follows:

(i) Horizontal component

$v_x=u \ cos \ \theta⁡$

where u is the speed of the particle at the moment when it passes through slit A. Since there is no force acting on the particle in the horizontal direction,

$v_x= \ constant $

(ii) Vertical component

$v_y=u_y+at$

$v_y=u \ sin⁡ \ \theta+at$

(iii)  Refer to Figure 17.53(b). The magnitude of the instantaneous velocity is given by

$v=\sqrt{v_x^2+v_y^2}$

The direction of v is given by

$tan \ \phi=\frac{v_y}{v_x}$ 

(d) Time taken to emerge from field

Let the distance between slit A and slit B be L. Then the time T taken by the particle to travel from A to B is given by

$T=\frac{L}{u \ cos \ \theta}$

EXAMPLE 17.28

Figure 17.54 shows a pair of parallel charged metal plates A and B. The uniform electric field existing in the space between the plates has magnitude $E=20 \ kV \ m^{-1}$. An electron travelling at constant speed $1.3 \times 10^7 \ m \ s^{-1}$enters the field through slit P and emerges through slit Q in plate B. The time taken by the electron to travel from P to Q is 5.0 ns. Determine
(a) the acceleration of the electron in the field
(b) the angle θ between plate B and the velocity before the electron enters the field
(c) the distance between P and Q

$(e/m_e =-1.76 \times 10^{11} C \ kg^{-1}$

Answer

(a) $a=\left(\frac{e}{m}\right)E$

$a=(1.76×10^11)(20×10^3)$

$a=3.52 \times 10^{15} \ m \ s^{-2}$

(b) Consider motion which is vertical to the plates.

$s_y=u_y t+\frac{1}{2} at^2$

If the electron starts from P and arrives at Q, then $s_y=0$. Given $t=5.0 \ ns$, $u=1.3 \times 10^7 \ m \ s^{-1}$

$0=(u \ sin \ \theta)t+\frac{1}{2} at^2$

$sin⁡ \ \theta=\frac{(3.52 \times 10^{15})(5.0 \times 10^{-9})}{2(1.3\times 10^7)}=0.6769$

$\theta= sin^{-1}(0.6769)=43^0$ to the plate 

(c)Motion parallel to the plates

$PQ=(u \ cos \ \theta)t$

$PQ=(1.3 \times 10^7)(cos⁡ \ 43^0)(5.0 \times 10^{-9})$

$PQ=4.8 \times 10^{-2} \ m$

17.6.4 Conservation of Energy of Charged Particle Moving in Electric Field

1. Suppose that at a particular instant a positively charged particle moving at speed u is at point A in an electric field, as shown in Figure 17.56. A short while later it arrives at point B moving at speed v. A p.d. of $\Delta V_AB$ volt exists across A and B.

2. Since the charged particle is in an electric field, an electric force has to act on it. As a result, work $W_AB$ is done by the electric force when the particle moves from A to B. We have

$W_{AB}=-\Delta U_{AB}$

where $\Delta U_{AB}$ is the change in electric potential energy of the charge system brought about by the work done by the electric force.

3. The change in electric potential energy is given by

$\Delta U_{AB}=q \Delta V_{AB}$

where q is the charge carried by the particle. Hence we have

$W_{AB}=-q\Delta V_{AB}$

According to the work–energy theorem, we have

$W_{AB}=\Delta K_{AB}$

or

$\Delta U_{AB}=-\Delta K_{AB}$

Where $\Delta K_{AB}$ is the change in kinetic energy of the particle when it travels from A to B. Combining the lats two expressions, we get

$-q \Delta V_{AB}=\Delta K_{AB}$

$-q(V_B-V_A)=K_B-K_A$

$K_A+qV_A=K_B+qV_B$

In other words

$(KE + electric \ PE) \ at A = (KE + electric \ PE) \ at B$

Hence, the total energy at A is equal to the energy at B (provided that the friction force experienced by the moving particle is negligible). This statement merely states the law of conservation of energy.

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