Question

has -22 a mass Q2) An alpha particle, the nucleus of a helium atom, of 6.64*10* kg and a charge of tae. What magni lude and direction of the electric field that will balance its weight? are​

Answer 1

Answer:

Vertical equilibrium of forces leads to the equality

$q | \overrightarrow{E}| = mg \\ \\ \implies \: | \overrightarrow{E}| = \frac{mg}{2e}$

Substituting the values given in the problem,we obtain

$| \overrightarrow{E}| = \frac{mg}{2e} \\ \\ = \frac{(6.64 \times {10}^{ - 27}kg)(9.8m \s ^{2} ) }{2(1.6 \times {10}^{ - 19}c } ) \\ \\ = 2.03 \times {10}^{ - 7} N/C$

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Answer 2

Answer:

Vertical equilibrium of forces leads to the equality

\begin{gathered}q | \overrightarrow{E}| = mg \\ \\ \implies \: | \overrightarrow{E}| = \frac{mg}{2e} \end{gathered}

q∣

E

∣=mg

⟹∣

E

∣=

2e

mg

Substituting the values given in the problem,we obtain

\begin{gathered} | \overrightarrow{E}| = \frac{mg}{2e} \\ \\ = \frac{(6.64 \times {10}^{ - 27}kg)(9.8m \s ^{2} ) }{2(1.6 \times {10}^{ - 19}c } ) \\ \\ = 2.03 \times {10}^{ - 7} N/C\end{gathered}

Explanation:

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