Showing posts with label Distance. Show all posts
Showing posts with label Distance. Show all posts

What distance must an electron move in a uniform potential gradient $200\;V/cm$ in order to gain kinetic energy $3.2 * 10 ^{-18}J$?


Given,
Potential Gradient $(E)$ = $200 V/cm$ = $200 * 10^3 \;V/m$
Kinetic Energy $(E_k)$
Charge of Electron $(q)$ = $1.6 * 10^{-19}\; C$
Mass of Electron $(m)$ = $9.1 * 10 ^{-31}\;kg$
Distance $(S)$ = ?

We know that,
$F = m.a$
 ⇒ $a$   =  $ \frac{F}{m}$  =  $\frac{q.E}{m}$   =   $\frac{1.6 * 10^{-19}\; * \; 200 * 10^3}{9.1 * 10 ^{-31}}$   =   $ 3.5 * 10^{15}\; m/s^2$ 
Now,
 $E_K = $$\frac{1}{2}$$m.v^2$
⇒ $v^2 = $$\frac{2. E_k}{m}$   =   $\frac{2 \; * \; (3.2 * 10^{-18})}{(9.1 * 10^{-31})}$    =   $7 * 10^{12}\;m/s$

Again, to Calculate the distance,
$v^2 = u^2 + 2as$                        [∵ $u = 0$]
$S = $$\frac{v^2}{2a}$   =   $\frac{7 * 10^{12}}{2 * (3.5 * 10^{15})}$   =   $0.001\;m$
Hence,
The required distance is $0.001\; m$.


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M5. Coordinates in space:










We consider $3-$ mutually perpendicular lines intersecting at a fixed point. Where this fixed point is called origin $(0)$ and the three line is called coordinate axis labeled as $x-$axis, $y-$axis and $z-$axis respectively as shown in figure.
1) Distance between two points:
Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be the two given points. Then the distance between two points $PQ$ is given by,
$PQ$ = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ .......... (i)
2) Section Formula:
Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be any points in the $3-$ dimensional space. Suppose a point $R(x, y, z)$ divides the join of $P$ and $Q$ in the ratio $m:n$. Then,
(i) Internal division:
The coordinates of $R(x, y, z)$ are 
$R(x, y, z)$ = $\left ( \frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}, \frac{mz_2+nz_1}{m+n} \right )$ .......... (ii)
(ii) External division:
The coordinates of $R(x, y, z)$ are 
$R(x, y, z)$ = $\left ( \frac{mx_2-nx_1}{m-n}, \frac{my_2-ny_1}{m-n}, \frac{mz_2-nz_1}{m-n} \right )$ .......... (iii)
3) Projection:
Projection of a line joining two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ on another line with direction cosine $l, m, n$. Then the direction cosines of $PQ$ are
$\frac{x_2-x_1}{r}, \frac{y_2-y_1}{r}, \frac{z_2-z_1}{r}$ .......... (iv)
Where, $PQ = r = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
If $\theta$ be the angle between $PQ$ and $AB$ then, 
      $cos\theta = \frac{x_2-x_1}{r}.l + \frac{y_2-y_1}{r}.m + \frac{z_2-z_1}{r}.n $ .......... (v)
Also the Projection of $PQ$ on $AB$ = $MN$ = $PQcos\theta$
$PQcos\theta = (x_2-x_1).l + (y_2-y_1).m + (z_2-z_1).n $ .......... (vi)

4) Direction Cosine of a line $(l, m, n)$:
If a line $AB$ makes an angle $\alpha, \beta, \gamma$ with the direction $x-$axis, $y-$axis and $z-$axis respectively. Then, $cos\alpha$, $cos\beta$  and  $cos\gamma$ are called direction cosines (d.c.s) of a line $AB$. It is generally denoted by $l, m, n$.
$l^2 + m^2 + n^2 = 1$ .......... (vii)
[$Cos\alpha,\;cos\beta,\;cos\gamma$   i.e.   $l,\;m,\;n$ are the direction cosines of the line $AB$.]

5) Direction Ratio $(a, b, c)$:
If a set of three numbers $a$, $b$, and $c$ proportional to the directions cosines $l$, $m$, and $n$ of a line, they are called its direction ratios. We have,
$\frac{l}{a}$ = $\frac{m}{b}$ = $\frac{n}{c}$ = $\frac{\sqrt{l^2 + m^2 + n^2}}{\sqrt{a^2 + b^2 + c^2}}$ .......... (viii)
           Hence,
$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}$,    $m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}$,    $n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$ .......... (ix)

6) Angle between two lines:
From figure, let, $(l_1,\;m_1,\;n_1)$ and $(l_2,\;m_2,\;n_2)$ be the direction cosines of two given lines $AB$ and $CD$ respectively.
Let $OP$ and $OQ$ be the lines through the origin $O$ parallel to the line $AB$ and $CD$ respectively.
To find the angle between two lines with direction cosine $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$,
$Cos\theta$ = $l_1l_2 + m_1m_2 + n_1n_2$ .......... (x)
(i) Condition of perpendicular of two lines: 
$l_1l_2 + m_1m_2 + n_1n_2 = cos\;90^0 = 0$ .......... (xi)
(ii) Condition of parallelism of two lines: 
$\frac{l_1}{l_2}$ = $\frac{m_1}{m_2}$ = $\frac{n_1}{n_2}$ = $\frac{\sqrt{l_1\,^2 + m_1\,^2 + n_1\,^2}}{\sqrt{l_2\,^2 + m_2\,^2 + n_2\,^2}}$ 
∴ $l_1 = l_2$;  $m_1 = m_2$,   $n_1 = n_2$ .......... (xii)
Which is the required condition of parallelism of two straight lines.

» Angle between two straight lines whose direction ratios are given:
Let $a_1, b_1, c_1$ and $a_2, b_2, c_2$ be the direction ratios of two lines whose corresponding direction cosines are $l_1, m_1, n_1$ and $l_2, m_2, n_2$. Then,
$l_1 = \frac{a_1}{\sqrt{a_1\,^2 + b_1\,^2 + c_1\,^2}}$,  $m_1 = \frac{b_1}{\sqrt{a_1\,^2 + b_1\,^2 + c_1\,^2}}$,  $n_1 = \frac{c_1}{\sqrt{a_1\,^2 + b_1\,^2 + c_1\,^2}}$
And similarly for $l_2, m_2, n_2$.
» If $\theta$ be the angle between the two lines, then
$cos\theta = l_1l_2 + m_1m_2 + n_1n_2$ = $\frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1\,^2 + b_1\,^2 + c_1\,2}\;\sqrt{a_2\,^2 + b_2\,^2 + c_2\,2}}$ .......... (xiii)

- If two lines are perpendicular,       $a_1a_2 + b_1b_2 + c_1c_2 = 0$ 
- Two lines will be parallel if          $\frac{a_1}{a_2}$ = $\frac{b_1}{b_2}$ = $\frac{c_1}{c_2}$
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E1.3 Kinematics (Distance and Displacement):

Motion of an object is defined as the change in position of the object with respect to its surroundings. If the object does not change their position with respect to their surroundings, which is said to be at rest.

Distance: Distance is a scalar quantity. The length of actual path followed by an object between its initial and final position is called distance. There is no directional component to a distance measurement. 


Displacement: Displacement is a vector quantity. It is independent of actual path followed by an object but depends only on the distance between initial and final position.
This is a best example to visualize what's the difference between Distance and Dispalcement.


Calculating Distance and Displacement: 

Here I want to illustrate the concept of distance and displacement.
From the Figure, If I walk $3m$ west and turn to $4m$ north. Then easily to calculate the distance what I traveled. 
i.e. $3m$ + $4m$ = $7m$. 
This $7m$ is just only a distance what I traveled and no need to explain the direction of travel. 

But, to determine Displacement; first of all I can draw an arrow where I started walking and extends to the point where I stopped walking.
From the Pythagorean Theorem: $a{^2}$ + $b{^2}$ = $c{^2}$. Since $c^2$ is our displacement squared; it should be written as: $(\Delta S){^2}$. 
⟹ $(\Delta S)$ = $\sqrt{3^2 + 4^2}$ = $\sqrt{9 + 16}$ = $\sqrt{25}$ = $5$
So, the displacement is $5m$ north west from the initial position. Where need to specify the direction to explain the displacement.

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