Capacitor's and Dielectrics | PPT Slide

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A potentiometer is 10 m long. It has a resistance of 20 Ω. It is connected in series with a battery of 3 V and a resistance of 10 Ω. What is the potential gradient along with wire?

Given,

Length of potentiometer wire $(l) = 10\;m$

Resistance of potentiometer $(R_p) = 20\; Ω$

Potential of battery $(V_b) = 3V$

Resistance of battery $(R_b) = 10 \; Ω$

Potential Gradient $(\frac{V}{l}) = \;?$

Let $I$ be the current flowing through the circuit.

Then, from Ohm's law,

$V_b = IR$     [where R= Total resistance of the circuit = $R_b + R_p$

or, $3 = I\;*\;(10+20)$

⇒ I = 0.1 A

Since the combination is in series, the current (I) all over the circuit is same.

So, p.d in AB $(V_{AB})$ = I * $R_{b}$ = 0.1 * 20 = 2 V

Now, Potential Gradient $(\frac{V}{l}) = \frac{2}{10}$ = 0.2V/m

Hence, the potential gradient along the wire is 0.2 V/m.

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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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