Answer :

Answer:

acceleration = 6.15 × 10¹⁴ m/s²

Explanation:

We can find the magnitude of the acceleration of an electron in a  3500 N/C electric field by using these formula:

[tex]\boxed{F_{electric}=E\cdot q}[/tex]

[tex]\boxed{F=m\cdot a}[/tex]

By combining both formulas, we come to this:

[tex]\boxed{E\cdot q=m\cdot a}[/tex]

where:

  • [tex]F[/tex] = force
  • [tex]E[/tex] = electric field
  • [tex]q[/tex] = electric charge
  • [tex]m[/tex] = mass
  • [tex]a[/tex] = acceleration

Given:

  • [tex]E[/tex] = 3500 N/C
  • [tex]q[/tex] (of an electron) = 1.602 × 10⁻¹⁹ C
  • [tex]m[/tex] (of an electron) = 9.11 × 10⁻³¹ kg

[tex]E\cdot q=m\cdot a[/tex]

[tex]3500\times 1.602\times10^{-19}=9.11\times10^{-31}\times a[/tex]

[tex]\displaystyle a=\frac{3.5\times10^3\times1.602\times10^{-19}}{9.11\times10^{-31}}[/tex]

[tex]\displaystyle a=\frac{3.5\times1.602}{9.11} \times10^{(3-19+31)}[/tex]

[tex]a=0.615\times10^{15[/tex]

[tex]\bf a=6.15\times10^{14}\ m/s^2[/tex]

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