Calculate the electrostatic force between two charges of \(\text{+6}\) \(\text{nC}\) and \(\text{+1}\)
\(\text{nC}\) if they are separated by a distance of \(\text{2}\) \(\text{mm}\).
\begin{align*}
F_e &= \frac{k Q_1 Q_2}{r^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{6} \times \text{10}^{-\text{9}})(\text{1} \times
\text{10}^{-\text{9}})}{(\text{2} \times \text{10}^{-\text{3}})^2} \\
&= \text{1,35} \times \text{10}^{-\text{2}}\text{ N}
\end{align*}
This force is repulsive since it is between two like charges.
What is the magnitude of the repulsive force between two pith balls (a pith ball is a small, light ball
that can easily be charged) that are \(\text{8}\) \(\text{cm}\) apart and have equal charges of
\(-\text{30}\) \(\text{nC}\)?
\begin{align*}
F_e &= \frac{k Q_1 Q_2}{r^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{30} \times \text{10}^{-\text{9}})(\text{30}
\times \text{10}^{-\text{9}})}{(\text{8} \times \text{10}^{-\text{2}})^2} \\
&= \text{1,27} \times \text{10}^{-\text{3}}\text{ N}
\end{align*}
This force is repulsive since it is between two like charges.
How strong is the attractive force between a glass rod with a \(\text{0,7}\) \(\mu\)C charge and a silk
cloth with a \(-\text{0,6}\) \(\mu\)C charge, which are \(\text{12}\) \(\text{cm}\) apart, using the
approximation that they act like point charges?
\begin{align*}
F_e &= \frac{k Q_1 Q_2}{r^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{0,6} \times \text{10}^{-\text{6}})(\text{0,7}
\times \text{10}^{-\text{6}})}{(\text{12} \times \text{10}^{-\text{2}})^2} \\
&= \text{0,26}\text{ N}
\end{align*}
This force is attractive since it is between two unlike charges.
Two point charges exert a \(\text{5}\) \(\text{N}\) force on each other. What will the resulting force be
if the distance between them is increased by a factor of three?
Let the charges be \(Q_{1}\) and \(Q_{2}\). For the first situation we have:
\[F_{e1} = \frac{k Q_1 Q_2}{r^2}\]
Now we increase the distance by a factor of three so we get:
\[F_{e2} = \frac{k Q_1 Q_2}{9r^2}\]
We now note the following:
\[(F_{e1})(r^2) = k Q_1 Q_2\]
and:
\[(F_{e2})(9r^2) = k Q_1 Q_2\]
Therefore:
\begin{align*}
(F_{e1})(r^2) & = (F_{e2})(9r^2) \\
F_{e1} & = (F_{e2})(9) \\
\therefore F_{e2} & = \text{0,56}\text{ N}
\end{align*}
Two point charges are brought closer together, increasing the force between them by a factor of 25. By
what factor was their separation decreased?
Let the charges be \(Q_{1}\) and \(Q_{2}\). For the first situation we have:
\[F_{e1} = \frac{k Q_1 Q_2}{r^2}\]
Now we decrease by an unknown factor, \(x\), the distance and we get:
\[F_{e2} = \frac{k Q_1 Q_2}{(xr)^2}\]
We now note the following:
and:
\[(F_{e2}) = \text{25}\frac{k Q_1 Q_2}{r^2}\]
Therefore:
\begin{align*}
\text{25}(\frac{k Q_1 Q_2}{r^2}) & = \frac{k Q_1 Q_2}{(xr)^2} \\
\text{25} & = \frac{1}{x^2} \\
\therefore x & = \text{0,2}
\end{align*}
If two equal charges each of \(\text{1}\) \(\text{C}\) each are separated in air by a distance of
\(\text{1}\) \(\text{km}\), what is the magnitude of the force acting between them?
\begin{align*}
F_e &= \frac{k Q_1 Q_2}{r^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{1})(\text{1})}{(\text{1} \times
\text{10}^{\text{3}})^2} \\
&= \text{9 000}\text{ N}
\end{align*}
This force is repulsive since it is between two like charges.
You should notice that even at a distance as large as \(\text{1}\) \(\text{km}\), the repulsive force is
substantial because \(\text{1}\) \(\text{C}\) is a very significant amount of charge.
Calculate the distance between two charges of \(\text{+4}\) \(\text{nC}\) and \(-\text{3}\) \(\text{nC}\)
if the electrostatic force between them is \(\text{0,005}\) \(\text{N}\).
\begin{align*}
F_e &= \frac{k Q_1 Q_2}{r^2} \\
\text{0,005} &= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{4} \times
\text{10}^{-\text{9}})(\text{3} \times \text{10}^{-\text{9}})}{(d)^2} \\
\text{0,005}d^{2} &= \text{1,07} \times \text{10}^{-\text{9}} \\
d^{2} &= \text{2,1576} \times \text{10}^{-\text{5}} \\
&= \text{4,6} \times \text{10}^{-\text{3}}\text{ m}
\end{align*}
For the charge configuration shown, calculate the resultant force on \(Q_2\) if:
- \(Q_1 = \text{2,3} \times \text{10}^{-\text{7}}\) \(\text{C}\)
- \(Q_2 = \text{4} \times \text{10}^{-\text{6}}\) \(\text{C}\)
- \(Q_3 = \text{3,3} \times \text{10}^{-\text{7}}\) \(\text{C}\)
- \(r_1 = \text{2,5} \times \text{10}^{-\text{1}}\) \(\text{m}\)
- \(r_2 = \text{3,7} \times \text{10}^{-\text{2}}\) \(\text{m}\)
We first calculate the force on \(Q_{2}\) from \(Q_{3}\):
\begin{align*}
F_{23} &= \frac{k Q_2 Q_3}{(r_{2})^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{4}
\times \text{10}^{-\text{6}})(\text{3,3} \times \text{10}^{-\text{7}})}{(\text{3,7} \times
\text{10}^{-\text{2}})^{2}} \\
&= \text{8,678}\text{ N}
\end{align*}
And then we calculate the force of \(Q_{1}\) on \(Q_{2}\). Note that for this force we must add \(r_{1}\)
and \(r_{2}\).
\begin{align*}
F_{21} &= \frac{k Q_1 Q_2}{(r_{1} + r_{2})^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{2,3}
\times \text{10}^{-\text{7}})(\text{4} \times \text{10}^{-\text{6}})}{(\text{2,5} \times
\text{10}^{-\text{1}} + \text{3,7} \times
\text{10}^{-\text{2}})^2} \\
&= \text{0,101}\text{ N}
\end{align*}
Next we note that the force of \(Q_{3}\) on \(Q_{2}\) is repulsive and the force of \(Q_{1}\) on
\(Q_{2}\) is also repulsive. So these two forces act in the same direction (towards the right). The
resultant force is:
\begin{align*}
F_{eR} &= F_{23} + F_{21} \\
&= \text{8,678}\text{ N} + \text{0,101}\text{ N} \\
&= \text{8,779}\text{ N to the right}.
\end{align*}
For the charge configuration shown, calculate the charge on \(Q_3\) if the resultant force
on \(Q_2\) is \(\text{6,3} \times \text{10}^{-\text{1}}\) \(\text{N}\) to the right and:
- \(Q_1 = \text{4,36} \times \text{10}^{-\text{6}}\) \(\text{C}\)
- \(Q_2 = -\text{7} \times \text{10}^{-\text{7}}\) \(\text{C}\)
- \(r_1 = \text{1,85} \times \text{10}^{-\text{1}}\) \(\text{m}\)
- \(r_2 = \text{4,7} \times \text{10}^{-\text{2}}\) \(\text{m}\)
We are told that the resultant force is \(\text{6,3} \times \text{10}^{-\text{1}}\) \(\text{N}\) to the
right. Since the force of \(Q_{1}\) on \(Q_{2}\) is attractive, the force of \(Q_{3}\) on \(Q_{2}\) must
be repulsive to cause a resultant force to the right (if it was also attractive, the resultant force would
be to the left). So we know that \(Q_{3}\) must be negative.
We first calculate the force on \(Q_{2}\) from \(Q_{1}\):
\begin{align*}
F_{e1} &= \frac{k Q_1 Q_2}{r^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{4,36} \times \text{10}^{-\text{6}})(\text{7}
\times \text{10}^{-\text{7}})}{(\text{1,85} \times \text{10}^{-\text{1}} + \text{4,7} \times
\text{10}^{-\text{2}})^2} \\
&= \text{0,510}\ldots\text{ N}
\end{align*}
Next we use this and the resultant force to find the force on \(Q_{2}\) from \(Q_{3}\)
\begin{align*}
F_{eR} &= F_{e1} + F_{e2} \\
F_{e2} &= \text{6,3} \times \text{10}^{-\text{1}}\text{ N} - \text{0,510}\ldots\text{ N} \\
&= \text{0,119}\ldots\text{ N}
\end{align*}
And then we calculate the charge on \(Q_{3}\):
\begin{align*}
F_{e2} &= \frac{k Q_2 Q_3}{r^2} \\
\text{0,119}\ldots &= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{7} \times
\text{10}^{-\text{7}})(Q_{3})}{(\text{4,7} \times \text{10}^{-\text{2}})^2} \\
\text{2,64}\ldots \times \text{10}^{-\text{4}} &= (\text{6,3} \times \text{10}^{\text{3}})(Q_{3}) \\
Q_{3} &= \text{4,2} \times \text{10}^{-\text{8}}\text{ C}
\end{align*}
Calculate the resultant force on \(Q_1\) given this charge configuration:
We first calculate the force on \(Q_{1}\) from \(Q_{2}\):
\begin{align*}
F_e &= \frac{k Q_1 Q_2}{r^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{1} \times \text{10}^{-\text{9}})(\text{2} \times
\text{10}^{-\text{9}})}{(\text{0,07})^2} \\
&= \text{3,7} \times \text{10}^{-\text{6}}\text{ N}
\end{align*}
And then we calculate the force of \(Q_{3}\) on \(Q_{1}\):
\begin{align*}
F_e &= \frac{k Q_1 Q_3}{r^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{3} \times \text{10}^{-\text{9}})(\text{2} \times
\text{10}^{-\text{9}})}{(\text{0,04})^2} \\
&= \text{3,4} \times \text{10}^{-\text{5}}\text{ N}
\end{align*}
The magnitude of the resultant force acting on \(Q_{1}\) can be calculated from the forces using
Pythagoras' theorem because there are only two forces and they act in the \(x\)- and \(y\)-directions:
\begin{align*}
F_{R}^{2} & = F_{x}^{2} + F_{y}^{2} \\
F_{R} & = \sqrt{(\text{3,7} \times \text{10}^{-\text{6}})^{2} + (\text{3,4} \times
\text{10}^{-\text{5}})^{2}} \\
& = \text{3,42} \times \text{10}^{-\text{5}}\text{ N}
\end{align*}
We can find the angle using trigonometry:
\begin{align*}
\tan \theta_{R} & = \frac{y \text{-component}}{x \text{-component}} \\
& = \frac{\text{3,42} \times \text{10}^{-\text{5}}}{\text{3,7} \times \text{10}^{-\text{6}}}\\
& = \text{9,2432}\ldots \\
\theta_{R} & = \text{83,8}\text{°}
\end{align*}
The final resultant force acting on \(Q_1\) is \(\text{3,42} \times \text{10}^{-\text{5}}\) \(\text{N}\)
acting at an angle of \(\text{83,8}\)\(\text{°}\) to the negative \(x\)-axis.
Calculate the resultant force on \(Q_1\) given this charge configuration:
We first calculate the force on \(Q_{1}\) from \(Q_{2}\):
\begin{align*}
F_e &= \frac{k Q_1 Q_2}{r^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{1} \times \text{10}^{-\text{9}})(\text{9} \times
\text{10}^{-\text{9}})}{(\text{0,65})^2} \\
&= \text{1,92} \times \text{10}^{-\text{7}}\text{ N}
\end{align*}
And then we calculate the force of \(Q_{3}\) on \(Q_{1}\):
\begin{align*}
F_e &= \frac{k Q_1 Q_3}{r^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{3} \times \text{10}^{-\text{9}})(\text{9} \times
\text{10}^{-\text{9}})}{(\text{0,4})^2} \\
&= \text{1,52} \times \text{10}^{-\text{6}}\text{ N}
\end{align*}
The magnitude of the resultant force acting on \(Q_{1}\) can be calculated from the forces using
Pythagoras' theorem because there are only two forces and they act in the \(x\)- and \(y\)-directions:
\begin{align*}
F_{R}^{2} & = F_{x}^{2} + F_{y}^{2} \\
F_{R} & = \sqrt{(\text{1,92} \times \text{10}^{-\text{7}})^{2} + (\text{1,52} \times
\text{10}^{-\text{6}})^{2}} \\
& = \text{1,52} \times \text{10}^{-\text{6}}\text{ N}
\end{align*}
We can find the angle using trigonometry:
\begin{align*}
\tan \theta_{R} & = \frac{y \text{-component}}{x \text{-component}} \\
& = \frac{\text{1,52} \times \text{10}^{-\text{6}}}{\text{1,92} \times \text{10}^{-\text{7}}}\\
& = \text{7,91666}\ldots \\
\theta_{R} & = \text{82,80}\text{°}
\end{align*}
The final resultant force acting on \(Q_1\) is \(\text{1,52} \times \text{10}^{-\text{6}}\) \(\text{N}\)
acting at an angle of \(\text{82,80}\)\(\text{°}\) to the positive \(x\)-axis.
Calculate the resultant force on \(Q_2\) given this charge configuration:
We first calculate the force on \(Q_{2}\) from \(Q_{1}\):
\begin{align*}
F_e &= \frac{k Q_1 Q_2}{r^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{8} \times \text{10}^{-\text{9}})(\text{3} \times
\text{10}^{-\text{9}})}{(\text{0,05})^2} \\
&= \text{8,63} \times \text{10}^{-\text{5}}\text{ N}
\end{align*}
And then we calculate the force of \(Q_{3}\) on \(Q_{2}\):
\begin{align*}
F_e &= \frac{k Q_1 Q_3}{r^2} \\
&= \frac{(\text{9,0} \times \text{10}^{\text{9}})(\text{3} \times \text{10}^{-\text{9}})(\text{2} \times
\text{10}^{-\text{9}})}{(\text{0,03})^2} \\
&= \text{5,99} \times \text{10}^{-\text{5}}\text{ N}
\end{align*}
The magnitude of the resultant force acting on \(Q_{2}\) can be calculated from the forces using
Pythagoras' theorem because there are only two forces and they act in the \(x\)- and \(y\)-directions:
\begin{align*}
F_{R}^{2} & = F_{x}^{2} + F_{y}^{2} \\
F_{R} & = \sqrt{(\text{8,63} \times \text{10}^{-\text{5}})^{2} + (\text{5,99} \times
\text{10}^{-\text{5}})^{2}} \\
& = \text{1,05} \times \text{10}^{-\text{4}}\text{ N}
\end{align*}
We can find the angle using trigonometry:
\begin{align*}
\tan \theta_{R} & = \frac{y \text{-component}}{x \text{-component}} \\
& = \frac{\text{5,99} \times \text{10}^{-\text{5}}}{\text{8,63} \times \text{10}^{-\text{5}}}\\
& = \text{0,694}\ldots \\
\theta_{R} & = \text{34,76}\text{°}
\end{align*}
The final resultant force acting on \(Q_1\) is \(\text{1,05} \times \text{10}^{-\text{4}}\) \(\text{N}\)
acting at an angle of \(\text{34,76}\)\(\text{°}\) to the positive \(x\)-axis.
