Two positive point charges, of which one is triple the other, repel with magnitude force 2.7 N in a
vacuum when the distance between them is 10 cm. Determine the charge of the small value.
Coulomb Constant in a vacuum
\( k_e=9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}} \).
Problem data:
- Charge 1: q1 = Q;
- Charge 2: q2 = 3Q;
- Electric force of repulsion between the charges: FE = 2.7 N;
- Distance between the charges: d = 10 cm;
- Coulomb Constant in a vacuum: \( k_e=9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}} \).
Problem diagram:
The force
FE12 is the repulsion force on sphere 1 due to sphere 2, the
force
FE21 is the repulsion force on the sphere 2 due to sphere 1, the
magnitude of these forces is equal to the
FE (Figure 1).
>
Figure 1
Solution
First, we need to convert the distance given in centimeters (cm) to meters (m) used in the
International System of Units (
SI).
\[
\begin{gather}
d=10\;\mathrm{\cancel{cm}}\times\frac{1\;\mathrm m}{100\;\mathrm{\cancel{cm}}}=0,1\;\mathrm m=1\times10^{-1}\;\mathrm m
\end{gather}
\]
Applying
Coulomb's Law, the magnitude of electric force is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{\small E}=k_e\frac{|\;q_1\;||\;q_2\;|}{r^2}}
\end{gather}
\]
\[
\begin{gather}
F_{\small E}=k_e\frac{|\;Q\;||\;3Q\;|}{d^2}\\[5pt]
3Q^2=\frac{F_{\small E}d^2}{k_e}\\[5pt]
Q=\sqrt{\frac{F_{\small E}d^2}{3k_e}}\\[5pt]
Q=\sqrt{\frac{\left(2,7\;\mathrm N\right)\left(1\times 10^{-1}\;\mathrm m\right)^2}{3\left(9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}}\right)}}\\[5pt]
Q=\sqrt{\frac{\left(2,7\;\mathrm{\cancel N}\right)\left(1\times 10^{-2}\;\mathrm{\cancel{m^2}}\right)}{3\left(9\times 10^9\;\mathrm{\frac{\cancel N.\cancel{m^2}}{C^2}}\right)}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{Q=1\times 10^{-6}\;\mathrm{C}}
\end{gather}
\]
Note: If the value of the small charge is
q1 = 1 × 10−6 C,
the value of the greatest is
q2 = 3 × 1 × 10−6 = 3 × 10−6 C.