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Neko [114]
3 years ago
14

An ionized oxygen molecule (O2+) at point A has charge +e and moves at 1.24 ✕ 103 m/s in the positive x-direction. A constant el

ectric force in the negative x-direction slows the molecule to a stop at point B, a distance of 0.766 mm past A on the x-axis. Calculate the x-component of the electric field and the potential difference between points A and B. (The mass of an oxygen molecule is 5.31 ✕ 10−26 kg and the fundamental charge is e = 1.60 ✕ 10−19 C.)
Physics
1 answer:
Helga [31]3 years ago
8 0

Answer:

\mathbf{E =3.33 \times 10^2 \ N/C}

\mathbf{ V_A - V_B = 0.2551 \ Volts}

Explanation:

Given that:

The charge on the ionized oxygen molecule = +e

The speed of the ionized oxygen molecule with this charge is 1.24 × 10³ m/s

distance travelled by the particle before rest is d = 0.766 m

According to the third equation of motion.

v^2 = u^2 +2as

v^2 = u^2 +2(\dfrac{-eE}{m}) s

0^2= u^2 +2(\dfrac{-eE}{m}) s

E = \dfrac{mu^2}{2e* \ s}

E = \dfrac{5.31 *10^{-26}* (1.24*10^3)^2}{2*1.6*10^{-19}*0.766*10^{-3}}

\mathbf{E =3.33 \times 10^2 \ N/C}

Thus, the electric field shows to be in the negative x-direction.

The potential difference between point A and B now is:

\Delta V = E.d \\ \\ V_A - V_B = 3.33 \times 10^2 \times 0.766 \times 10^{-3}

\mathbf{ V_A - V_B = 0.2551 \ Volts}

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{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

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{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

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• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

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{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

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{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

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