Ask & Answer Question

Ncert Solutions Physics Class 12th Physics Ncert Solutions Class 12th Physics Class 12th Electric Charges and Fields

1.33 A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed $v_{x}$ (like particle 1 in Fig. 1.33). The length of plate is L and a uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $\frac{q E L^{2}}{2 m v_{x}}$

Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

6 Views | Posted 5 months ago

Asked by Shiksha User

2 Answers

Answered by

alok kumar singh | Contributor-Level 10

5 months ago

1.33 Charge on a particle of mass m = -q

Velocity of the particle = $v_{x}$

Length of the plates = L

Magnitude of the uniform electric field between the plates = E

Mechanical force, F = Mass (m) $*$ Acceleration (a)

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

a = $\frac{q E}{m}$ ………(1) { as electric force F = qE }

Time taken by the particle to cross the field of length L is given by.

t = $\frac{L e n g t h o f t h e p l a t e}{V e l o c i t y o f t h e p a r t i c l e}$ = $\frac{L}{v_{x}}$ ………….(2)

In the vertical direction, initial velocity, u = 0

From the relation s = ut + $\frac{1}{2}$ a $t^{2}$

We get, s = 0 + $\frac{1}{2} \frac{q E}{m} ({\frac{L}{v_{x}})}^{2}$

s = $\frac{q e L^{2}}{2 m {v_{x}}^{2}}$

Hence, vertical deflection of the particle at the far edge of the plate is $\frac{q e L^{2}}{2 m {v_{x}}^{2}}$ . This is similar to the motion of horizontal projectiles under gravity.

0 Upvotes 0 Downvotes

Answered by

alok kumar singh | Contributor-Level 10

5 months ago

1.33 Charge on a particle of mass m = -q

Velocity of the particle = $v_{x}$

Length of the plates = L

Magnitude of the uniform electric field between the plates = E

Mechanical force, F = Mass (m) $\times$ Acceleration (a)

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

a = $\frac{q E}{m}$ ………(1) { as electric force F = qE }

Time taken by the particle to cross the field of length L is given by.

t = $\frac{L e n g t h o f t h e p l a t e}{V e l o c i t y o f t h e p a r t i c l e}$ = $\frac{L}{v_{x}}$ ………….(2)

In the vertical direction, initial velocity, u = 0

From the relation s = ut + $\frac{1}{2}$ a $t^{2}$

We get, s = 0 + $\frac{1}{2} \frac{q E}{m} ({\frac{L}{v_{x}})}^{2}$

s = $\frac{q e L^{2}}{2 m {v_{x}}^{2}}$

Hence, vertical deflection of the particle at the far edge of the plate is $\frac{q e L^{2}}{2 m {v_{x}}^{2}}$ . This is similar to the motion of horizontal projectiles u

...more

1.33 Charge on a particle of mass m = -q

Velocity of the particle = $v_{x}$

Length of the plates = L

Magnitude of the uniform electric field between the plates = E

Mechanical force, F = Mass (m) $\times$ Acceleration (a)

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

a = $\frac{q E}{m}$ ………(1) { as electric force F = qE }

Time taken by the particle to cross the field of length L is given by.

t = $\frac{L e n g t h o f t h e p l a t e}{V e l o c i t y o f t h e p a r t i c l e}$ = $\frac{L}{v_{x}}$ ………….(2)

In the vertical direction, initial velocity, u = 0

From the relation s = ut + $\frac{1}{2}$ a $t^{2}$

We get, s = 0 + $\frac{1}{2} \frac{q E}{m} ({\frac{L}{v_{x}})}^{2}$

s = $\frac{q e L^{2}}{2 m {v_{x}}^{2}}$

Hence, vertical deflection of the particle at the far edge of the plate is $\frac{q e L^{2}}{2 m {v_{x}}^{2}}$ . This is similar to the motion of horizontal projectiles under gravity.

less

0 Upvotes 0 Downvotes

2 Answers

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question