A very small ball has a mass of 5.00 × 10⁻³kg and a charge of 4.00μC. What magnitude electric field directed upward will balance the weight of the ball so that the ball is suspended motionless above the ground?

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]