Suppose the electron enters the electric field at time t - 0- What is the velocity v- of the electron at time t1 when it leaves the plates-

The correct option is B v⃗ = v_0 î + (eE_yL_1mv_0) ĵ Force on electron =F⃗=qE⃗ F⃗=( e)( E_yĵ) F⃗=eE_yĵ Acceleration =a⃗=F⃗m a⃗=eE_ym ...

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Byju's Answer

Standard XII

Physics

A Mysterious Force

Suppose the e...

Question

Suppose the electron enters the electric field at time $t = 0$ . What is the velocity $\to v$ of the electron at time $t_{1}$ when it leaves the plates?

\to v = (\frac{e E_{y} L_{1}}{m v_{0}})^i + v_{0}^j

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\to v = v_{0}^i + (\frac{e E_{y} L_{1}}{m v_{0}})^j

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\to v = v_{0}^i - (\frac{e E_{y} L_{1}}{m v_{0}})^j

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\to v = - (\frac{e E_{y} L_{1}}{m v_{0}})^i + v_{0}^j

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Solution

The correct option is B $\to v = v_{0}^i + (\frac{e E_{y} L_{1}}{m v_{0}})^j$
Force on electron $= \to F = q \to E$

$⟹ \to F = (- e) (- E_{y}^j)$

$⟹ \to F = e E_{y}^j$

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

$⟹ \to a = \frac{e E_{y}}{m}^j$

Now, initial speed $= \to u = v_{o}^i$

Since there is no force along X-axis hence $v_{x}$ does not changes.

Hence, time taken to cross plate along X-direction $= t = \frac{L_{1}}{v_{x}} = \frac{L_{1}}{v_{o}}$

final velocity= $\to v = \to u + \to a t$

$\to v = v_{o}^i + (\frac{e E_{y} L_{1}}{m v_{o}})^j$

Answer-(B)

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Similar questions

Q. Assertion :If

\to v =^i +^j +^k

and

\to b = 2 \to i + 2 \to j

, then

(P r o j_{b} v) =^i +^j

and component of

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orthogonal to

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Reason:

P r o j_{b} v = \frac{\to v \cdot \to b}{{∣ ∣ \to b ∣ ∣}^{2}} \to b

and component of

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orthogonal to

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and

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Suppose the electron enters the electric field at time t=0. What is the velocity →v of the electron at time t1 when it leaves the plates?

Suppose the electron enters the electric field at time $t = 0$ . What is the velocity $\to v$ of the electron at time $t_{1}$ when it leaves the plates?