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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 electric 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.)

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Answer:

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

[tex]\mathbf{ V_A - V_B = 0.2551 \ Volts}[/tex]

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.

[tex]v^2 = u^2 +2as[/tex]

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

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

[tex]E = \dfrac{mu^2}{2e* \ s}[/tex]

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

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

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

The potential difference between point A and B now is:

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

[tex]\mathbf{ V_A - V_B = 0.2551 \ Volts}[/tex]

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