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### Motion Of Charge In Uniform Electric Field

Learn formula of motion of a charged particle in a uniform electric field, Practice calculation of maximum height of charge projected in uniform electric field.

# Parallel and Anti-Parallel Motion of Charge in a Uniform Electric Field

• Consider a positive charge in a uniform electric field, as shown in figure.

### Case I:   Initial velocity of particle in the direction of field vector

When angle between  $$\vec u$$ and $$\vec E$$  is $$0°$$

$$\left ( \vec u || \vec E \right)$$

• Since, velocity and acceleration both act along the same line so, the particle will travel along a straight line.

### Case II:  Initial velocity of particle is opposite to direction of field vector

When angle between $$\vec u$$ and $$\vec E$$ is $$180°$$

• Since, velocity and acceleration both act along the same line so, the particle will travel along a straight line.

Note: Force on a negative charge acts in opposite direction of the field.

#### A particle of mass $$m=1 kg$$ and charge $$q = 1\,C$$ is projected with zero initial velocity $$(u)$$ within an electric field of $$\vec E = 10^5 \,V/m$$ in the direction of field vector. Find the velocity$$(v)$$ of charge after covering a distance $$s = 20\,m$$.

A $$2\times 10^5\,m/sec$$

B $$2\times 10^3\,m/sec$$

C $$3\times 10^8\,m/sec$$

D $$0$$

×

Direction of acceleration is as shown in figure.

Using, $$v^2=u^2+2as, u = 0$$

$$v^2 = 0^2+2 \left ( \dfrac {q\vec E}{m}\right)(s)$$              $$\left (\because a= \dfrac {q\vec E}{m}\right)$$

where s = distance

Given: $$q=1\,C,\,\,E=10^5\,V/m,\,\,m=1\,kg\,,\,\,s=20\,m$$

$$v^2 = 2\left(\dfrac {10^5\times20}{1}\right)$$

$$v$$ = $$2\times10^3\, m/sec$$

### A particle of mass $$m=1 kg$$ and charge $$q = 1\,C$$ is projected with zero initial velocity $$(u)$$ within an electric field of $$\vec E = 10^5 \,V/m$$ in the direction of field vector. Find the velocity$$(v)$$ of charge after covering a distance $$s = 20\,m$$.

A

$$2\times 10^5\,m/sec$$

.

B

$$2\times 10^3\,m/sec$$

C

$$3\times 10^8\,m/sec$$

D

$$0$$

Option B is Correct

# Motion of Charge in a Uniform Electric Field

• A charge placed in an electric field experiences a force. This force causes an acceleration.
• Path of particle can be determined by signs of its velocity and acceleration.

Case I

• The path of particle will be a straight line when velocity and acceleration of the particle are collinear.

Conclusion: Positive charge experiences the force in the direction of electric field and negative charge experiences the force opposite to the direction of electric field.

Case II

• When the particle with initial velocity (u) makes an angle ($$\theta$$) with acceleration, the path of particle will be parabolic.

Path of positive particle in a uniform electric field $$[\;\vec u\;\nparallel\;\vec a\;]$$

Path of negative particle in a uniform electric field  $$[\;\vec u\;\nparallel\;\vec a\;]$$

Conclusion: Direction of electric field can be determined by interpreting the direction of velocity and acceleration when the path of particle is known.

#### Choose the direction of electric field responsible for the shown trajectory of a proton.

A

B

C

D

×

The direction of acceleration is in downward direction as shown in figure.

The path is parabolic in X-Y plane.So, net acceleration must be in negative Y-direction.

$$\vec a=\dfrac {q\vec E}{m}$$

Direction of $$\vec a$$= negative Y-direction

So, direction of $$\vec E=$$ negative Y-direction

### Choose the direction of electric field responsible for the shown trajectory of a proton.

A
B
C
D

Option C is Correct

# Vertical Projection of Charge in a Uniform Electric Field

• Consider a capacitor placed in a gravity free region where electric field is in downward direction, as shown in figure.

• On moving from negative plate towards positive plate, the positive charge experiences a force which produces constant acceleration in the direction of electric field.

• The particle will return without touching positive plate if the velocity of particle becomes zero just before approaching the positive plate.
• For this to happen,

$$d= \dfrac {u^2} {2 \left( \dfrac {q\vec E}{m} \right) }$$

• This case is similar to motion of particle under gravity.
• Maximum Height = $$\dfrac {u^2} {2g}$$

replacing $$g\rightarrow \dfrac {q\vec E}{m}$$

Maximum Height = $$\dfrac {u^2} {2 \left( \dfrac {q\vec E}{m} \right) }$$

#### A capacitor is placed in a gravity free region. Find the value of electric field $$\vec E$$ , so that the particle of charge q=1 C and mass m=1 kg returns to the point of projection without touching the other plate of the capacitor. Initial velocity u = 1 m/s, distance between the plates d = 0.5 mm.

A $$zero$$

B $$10\, V/m$$

C $$10^3 \,V/m$$

D $$10^5\, V/m$$

×

The direction of acceleration is downward direction, as shown in figure.

To return back to point of projection without touching other plate,

Maximum height $$= 0.5\,mm$$

a = acceleration

$$\dfrac {u^2}{2a}=d$$

$$\implies \dfrac {u^2}{2\vec a}= 0.5 mm$$

$$\implies \dfrac {(1)^2} {2 \left( \dfrac {q\vec E}{m} \right) } =0.5 mm$$

$$\implies \dfrac {m}{q\vec E}=1 mm$$

$$\implies\vec E=\dfrac {m}{q({1\,mm)}}$$

$$\implies\vec E=\dfrac {1}{1{(10^{-3})}}=10^3\,V/m$$

### A capacitor is placed in a gravity free region. Find the value of electric field $$\vec E$$ , so that the particle of charge q=1 C and mass m=1 kg returns to the point of projection without touching the other plate of the capacitor. Initial velocity u = 1 m/s, distance between the plates d = 0.5 mm.

A

$$zero$$

.

B

$$10\, V/m$$

C

$$10^3 \,V/m$$

D

$$10^5\, V/m$$

Option C is Correct

# Projectile Motion of Charge in a Uniform Electric Field

• Consider a capacitor which is placed in a gravity free region where electric field is in direction shown in figure.

• Particle of charge +q and mass m experiences a force in the direction of electric field when projected from negative plate with a speed  u m/sec making an angle $$\theta$$ with negative plate towards positive plate.
• This forces causes an acceleration,  $$\vec a=\dfrac {q\vec E}{m}$$

• In this case, particle will act like projectile.
• Replace  $$g=\dfrac {q\vec E}{m}$$

• To return just without touching the positive plate, maximum height attained by particle should be equal to the separation between plates of capacitor.

$$H_{max}=\dfrac {u^2sin^2\theta}{2 \left( \dfrac {q\vec E}{m}\right)}$$            $$\left(\because H_{max}=\dfrac {u^2sin^2\theta}{2g}\right)$$

$$d=\dfrac {u^2sin^2\theta}{2 \left( \dfrac {q\vec E}{m}\right)}$$

$$\vec E=\dfrac {mu^2sin^2\theta}{2qd}$$

#### A particle of charge q = 1 C and mass m=1 kg is projected from the negative plate of capacitor of plate separation d = 1 mm. If the initial velocity of particle is u = 2 m/sec and angle of projection $$\theta=60°$$, then find the value of E for which the particle return to negative plate just without touching the positive plate.

A $$zero$$

B $$1500\, V/m$$

C $$15\, V/m$$

D $$150\, V/m$$

×

The direction of acceleration of particle will be downward, as shown in figure.

Trajectory of particle is shown in figure.

To return without touching the positive plate maximum height of projectile = d, a= acceleration

$$d=\dfrac {u^2sin^2\theta}{2a}$$

$$\dfrac {u^2sin^2\theta}{2 \left( \dfrac {q\vec E}{m}\right)}=d$$

$$\vec E =\dfrac {u^2sin^2\theta\; m}{2 qd}$$

$$\vec E =\dfrac {(2)^2\times (3) \times1}{2\times1\times(4)\times 1\times10^{-3} }$$

$$\vec E=1.5\times10^3\, V/m$$

$$\vec E=1500\; \,V/m$$

### A particle of charge q = 1 C and mass m=1 kg is projected from the negative plate of capacitor of plate separation d = 1 mm. If the initial velocity of particle is u = 2 m/sec and angle of projection $$\theta=60°$$, then find the value of E for which the particle return to negative plate just without touching the positive plate.

A

$$zero$$

.

B

$$1500\, V/m$$

C

$$15\, V/m$$

D

$$150\, V/m$$

Option B is Correct

# Calculation of Maximum Distance Covered by the Particle before it Stops Instantaneously

• Consider a positive charge in a uniform electric field, as shown in figure.

## Case I: Initial velocity of particle in the direction of field vector

When angle between  $$\vec u$$ and $$\vec E$$  is $$0°$$

$$\left ( \vec u || \vec E \right)$$

• Since velocity and acceleration both act along the same line so, the particle will travel along a straight line.

## Case II: Initial velocity of particle is opposite to direction of field vector

When angle between $$\vec u$$ and $$\vec E$$ is $$180°$$

• Since, velocity and acceleration both act along the same line, the particle will travel along a straight line.

#### A particle of mass $$m=1kg$$ and charge $$q = –2C$$ is given a velocity of $$u=2\times10^3m/sec$$ in the direction of an electric field of intensity $$\vec E=3\times10^5V/m$$  within the field. Calculate the maximum distance $$(s)$$ covered before it stops instantaneously.

A $$0$$

B $$\dfrac {1}{3}m$$

C $$\dfrac {10}{3}m$$

D $$300 \,m$$

×

Direction of acceleration is as shown in figure.

Acceleration(a) of particle is given as

$$a=\dfrac {q\vec E}{m}$$$$=\dfrac {-2\times3\times10^5}{1}$$

$$a=-6\times 10^5m/sec^2$$

The particle will stop when v = 0.

Distance(s) covered till this point,

using, $$v^2=u^2+2as$$

$$s=\dfrac{v^2-u^2}{2a}$$

$$s=\dfrac {0^2-(2\times10^3)^2}{-2\times6\times10^5}$$

$$s=\dfrac {1}{3}\times10$$

$$s=\dfrac {10}{3}m$$

### A particle of mass $$m=1kg$$ and charge $$q = –2C$$ is given a velocity of $$u=2\times10^3m/sec$$ in the direction of an electric field of intensity $$\vec E=3\times10^5V/m$$  within the field. Calculate the maximum distance $$(s)$$ covered before it stops instantaneously.

A

$$0$$

.

B

$$\dfrac {1}{3}m$$

C

$$\dfrac {10}{3}m$$

D

$$300 \,m$$

Option C is Correct

# Calculation of Maximum Height of Charge Projected in Uniform Electric Field

• Consider a positively charged particle projected with a speed u at an angle  $$\theta$$  from horizontal where electric field is present in vertically downward direction.

• Force on charge particle is shown in figure.

• The particle will experience a constant force in direction of electric field causing an acceleration a,

$$\vec a = \dfrac {q\vec E}{m}$$  (downwards)

• This problem can be compared to a particle projected at an angle $$\theta$$ with horizontal under gravity.
• Maximum Height,

$$H_{max}=\dfrac {u^2sin^2\theta}{2g}$$

• Here, acceleration is as shown in figure.

• The maximum height attained by charge when it is projected in uniform electric field is as shown in figure.
• Maximum Height,

$$H_{max}=\dfrac{u^2sin^2\theta}{2\left(\dfrac{q\vec E}{m} \right)}$$

Conclusion: This problem can be considered as projectile motion under gravity by replacing acceleration due to gravity (g) with acceleration$$\left(\dfrac {q\vec E}{m}\right)$$.

Note: Here, acceleration due to gravity is negligible.

#### Calculate maximum height $$(H_{max})$$ attained by the particle when it is projected with a speed of  $$u=10\,m/sec$$  at an angle  $$\theta=30°$$  with horizontal in a gravity free region where electric field is present in a vertically downward direction. [$$q =+1\,C,$$ $$\vec E$$=$$10 \,V/m,$$ $$m = 1\,kg$$]

A $$2\,m$$

B $$3\,m$$

C $$2.5\,m$$

D $$1.25\,m$$

×

Trajectory of particle is shown in figure.

Maximum height of particle is given as

$$H_{max}=\dfrac {u^2sin^2\theta}{2a}$$           (here a = acceleration)

$$H_{max}=\dfrac {(10)^2\times(sin30°)^2}{2\times\left(\frac{1\times10}{1}\right)}$$

$$H_{max}=1.25\,m$$

### Calculate maximum height $$(H_{max})$$ attained by the particle when it is projected with a speed of  $$u=10\,m/sec$$  at an angle  $$\theta=30°$$  with horizontal in a gravity free region where electric field is present in a vertically downward direction. [$$q =+1\,C,$$ $$\vec E$$=$$10 \,V/m,$$ $$m = 1\,kg$$]

A

$$2\,m$$

.

B

$$3\,m$$

C

$$2.5\,m$$

D

$$1.25\,m$$

Option D is Correct