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Sagot :
Answer:
electric field = 1.225×10³ N/C
Explanation:
In order that the ball is suspended motionless above the ground, the resultant force, which comprised of weight and electrical force, has to equal to 0.
[tex]\begin{aligned}\Sigma F&=F_{elec}-w\\0&=F_{elec}-w\\\bf F_{elec}&= w\end{aligned}[/tex]
Since:
[tex]\boxed{F_{elec}=E\cdot q}[/tex]
and
[tex]\boxed{w=mg}[/tex]
where:
- [tex]F_{elec}=\texttt{electrical force}[/tex]
- [tex]E=\texttt{electric field}[/tex]
- [tex]q=\texttt{charge}[/tex]
- [tex]w=\texttt{weight}[/tex]
- [tex]m=\texttt{mass}[/tex]
- [tex]g=\texttt{gravity}=9.8\ m/s^2[/tex]
Given:
- [tex]m=5.00\times10^{-3}\ kg[/tex]
- [tex]q=4.00\ \mu C[/tex]
Then:
[tex]\begin{aligned}F_{elec}&=w\\ Eq&=mg\\E(4.00\times10^{-6})&=5.00\times10^{-3}\times9.8\\E&=(5.00\times9.8\div4.00)\times10^{-3+6}\\E&=12.25\times10^3\\E&=\bf1.225\times10^4\ N/C\end{aligned}[/tex]
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