Test Index

Electrostatics Part 1

Section: Physics

Question : 21 of 100

Marks: +1, -0

m

)

X

V_{0}.

\vec{E} = -E \hat{j}

x = d

x > d

Solution:

Let particle have charge

q

and mass 'm'Solve for

(q, m)

mathematically

F_{x} = 0, a_{x} = 0, (v)_{x} = \text{constant}

time taken to reach at 'P'

= \frac{d}{v_{0}} = t_{0} (\text{let})

...... (1)

(Along

- y), y_{0} = 0 + \frac{1}{2} \cdot \frac{qE}{m}

\cdot t_{0}^{2}

.... .(2)

v_{x} = v_{0}

v = u + at

(along -ve 'y')

speed

v_{y0} = \frac{qE}{m} \cdot t_{0}

\tan \theta = \frac{v_{y}}{v_{x}} = \frac{q E t_{0}}{m \cdot v_{0}},

\left(t_{0} = \frac{d}{v_{0}}\right)

\tan \theta = \frac{qEd}{m \cdot v_{0}^{2}}

\left[ \text{slope} = -\frac{qEd}{m v_{0}^{2}} \right]

Now we have to find eq

^{n}

of straight line

whose slope is

-\frac{qEd}{m v_{0}^{2}}

and it pass throughpoint

\rightarrow (d, -y_{0})

Because after

x > d

No electric field

\Rightarrow F_{\text{net}} = 0, \vec{v} = \text{const.}

y = mx + c, \left\{ \begin{array}{l} m = \frac{qEd}{m v_{0}^{2}} \\ (d, -y_{0}) \end{array} \right.

-y_{0} = -\frac{qEd}{m v_{0}^{2}} \cdot d + c

\Rightarrow c = -y_{0} + \frac{qEd^{2}}{m v_{0}^{2}}

Put the value

y = -\frac{qEd}{m v_{0}^{2}} x - y_{0} + \frac{qEd^{2}}{m v_{0}^{2}}

y_{0} = \frac{1}{2} \cdot \frac{qE}{m} \left(\frac{d}{v_{0}}\right)^{2}

= \frac{1}{2} \frac{qEd^{2}}{m v_{0}^{2}}

y = -\frac{qEdx}{m v_{0}^{2}} - \frac{1}{2} \frac{qEd^{2}}{m v_{0}^{2}} + \frac{qEd^{2}}{m v_{0}^{2}}

y = -\frac{qEd}{m v_{0}^{2}} x + \frac{1}{2} \frac{qEd^{2}}{m v_{0}^{2}}

\left[ y = \frac{qEd}{m V_{0}^{2}} \left(\frac{d}{2} - x\right) \right]

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