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Byju's Answer

Standard IX

Physics

Derivation of Position-Time Relation by Graphical Method

A uniform ele...

Question

A uniform electric field of strength exists in a region. An electron (charge $- e$ , mass $m$ ) enters a point $A$ with velocity $V^j$ . It moves through the electric field & exits at point $B$ . Then:

\to E = - \frac{2 a m v^{2}}{e d^{2}}^i

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Rate of work done by the electric field at

B

\frac{4 m a^{2} v^{3}}{d^{3}}

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Rate of work by the electric field at

A

is zero

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Velocity at

B

\frac{2 a v}{d}^i + v^j

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Solution

The correct options are
A $\to E = - \frac{2 a m v^{2}}{e d^{2}}^i$
B Rate of work done by the electric field at $B$ is $\frac{4 m a^{2} v^{3}}{d^{3}}$
D Velocity at $B$ is $\frac{2 a v}{d}^i + v^j$

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Similar questions

Q. A uniform electric field of strength E exists in a region. An electron (charge = –e and mass m) enters a point A(a, 0) with velocity v

^j

. It moves through the electric field and exits at point B(2a, d) with velocity

u^i + v^j

. Then:

Q. A uniform electric field of strength

E

exists in a region. An electron enters a point

A

with velocity

v

as shown. It moves through the field and reaches a point

B

. Velocity of particle at

B

2 v

30^{o}

with

x -

axis. Then,

Q. A particle of charge +q and mass m moving under the influence of a uniform electric field

E^i

and a uniform magnetic field

B^k

follows trajectory from P to Q as shown in figure. The velocities at P and Q are

v^i

and

- 2 v^j

respectively. Which of the following statement(s) is/are correct

Q. A particle of charge

q

and mass

m

moving under the influence of uniform electric field

E^i

and a uniform magnetic field

- B^i

moves from point

P (0, a)

Q (2 a, a)

where

O (0, 0)

is origin. The velocity of the particle at Point

P

v^i

and at

Q

2 v^i

. The rate of work done by the electric field & magnetic field on particle at point

P

will be respectively:

Q. A charged particle of mass

m

and charge

q

, moving under the influence of a uniform electric field

E^i

and a uniform magnetic field

B^k

, follows a trajectory from point

P

Q

as shown in figure. The velocities at

P

and

Q

are respectively,

v^i and - 2 v^j

. Then, which of the following statements (A, B, C, D) are correct? (Trajectory shown is schematic and not to scale)

(A)

E = \frac{3}{4} (\frac{m v^{2}}{q a})

(B)

Rate of work done by the electric field at

P

\frac{3}{4} (\frac{m v^{2}}{a})

(C)

Rate of work done by both the fields at

Q

is zero.

(D)

The difference between the magnitude of angular momentum of the particle at

P

and

Q

2 m a v

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A uniform electric field of strength exists in a region. An electron (charge −e, mass m) enters a point A with velocity V^j. It moves through the electric field & exits at point B. Then:

The correct options are A →E=−2amv2ed2^i B Rate of work done by the electric field at B is 4ma2v3d3 D Velocity at B is 2avd^i+v^j

A uniform electric field of strength exists in a region. An electron (charge $- e$ , mass $m$ ) enters a point $A$ with velocity $V^j$ . It moves through the electric field & exits at point $B$ . Then:

The correct options are
A $\to E = - \frac{2 a m v^{2}}{e d^{2}}^i$
B Rate of work done by the electric field at $B$ is $\frac{4 m a^{2} v^{3}}{d^{3}}$
D Velocity at $B$ is $\frac{2 a v}{d}^i + v^j$