More Examples on Coulomb's Law

Three point charges are in a straight line.Their charges are \(Q_1 = \text{+2} \times \text{10}^{-\text{9}}\text{ C}\), \(Q_2 = \text{+1} \times \text{10}^{-\text{9}}\text{ C}\) and \(Q_3 = -\text{3} \times \text{10}^{-\text{9}}\text{ C}\). The distance between \(Q_1\) and \(Q_2\) is \(\text{2} \times \text{10}^{-\text{2}}\) \(\text{m}\) and the distance between \(Q_2\) and \(Q_3\) is \(\text{4} \times \text{10}^{-\text{2}}\) \(\text{m}\). What is the net electrostatic force on \(Q_2\) due to the other two charges?

Step 1: Determine what is required

We need to calculate the net force on \({Q}_{2}\). This force is the sum of the two electrostatic forces - the forces between \({Q}_{1}\) on \({Q}_{2}\) and \({Q}_{3}\) on \({Q}_{2}\).

Step 2: Determine how to approach the problem

• We need to calculate the two electrostatic forces on \({Q}_{2}\), using Coulomb's law.

• We then need to add up the two forces using our rules for adding vector quantities, because force is a vector quantity.

Step 3: Determine what is given

We are given all the charges and all the distances.

Step 4: Calculate the magnitude of the forces.

Force on \(Q_2\) due to \(Q_1\):

\begin{align*} F_1 & = k\frac{Q_1Q_2}{r^2} \\ &= (\text{9.0} \times \text{10}^{\text{9}})\frac{(\text{2} \times \text{10}^{-\text{9}})(\text{1} \times \text{10}^{-\text{9}})}{(\text{2} \times \text{10}^{-\text{2}})^2} \\ &= (\text{9.0} \times \text{10}^{\text{9}})\frac{(\text{2} \times \text{10}^{-\text{9}})(\text{1} \times \text{10}^{-\text{9}})}{(\text{4} \times \text{10}^{-\text{4}})} \\ &= \text{4.5} \times \text{10}^{-\text{5}}\text{ N} \end{align*}

Force on \(Q_2\) due to \(Q_3\):

\begin{align*} F_3 & = k\frac{Q_2Q_3}{r^2} \\ &= (\text{9.0} \times \text{10}^{\text{9}})\frac{(\text{1} \times \text{10}^{-\text{9}})(\text{3} \times \text{10}^{-\text{9}})}{(\text{4} \times \text{10}^{-\text{2}})^2} \\ &= (\text{9.0} \times \text{10}^{\text{9}})\frac{(\text{1} \times \text{10}^{-\text{9}})(\text{3} \times \text{10}^{-\text{9}})}{(\text{16} \times \text{10}^{-\text{4}})} \\ &= \text{1.69} \times \text{10}^{-\text{5}}\text{ N} \end{align*}

Step 5: Vector addition of forces

We know the force magnitudes but we need to use the charges to determine whether the forces are repulsive or attractive. It is helpful to draw the force diagram to help determine the final direction of the net force on \(Q_2\).We choose the positive direction to be to the right (the positive \(x\)-direction).

The force between \(Q_1\) and \(Q_2\) is repulsive (like charges). This means that it pushes \(Q_2\) to the right, or in the positive direction.

The force between \(Q_2\) and \(Q_3\) is attractive (unlike charges) and pulls \(Q_2\) to the right.

Therefore both forces are acting in the positive direction.

Therefore,\begin{align*} F_R & = \text{4.5} \times \text{10}^{-\text{5}}\text{ N} + \text{1.69} \times \text{10}^{-\text{5}}\text{ N} \\ & = \text{6.19} \times \text{10}^{-\text{5}}\text{ N} \end{align*}

The resultant force acting on \(Q_2\) is \(\text{6.19} \times \text{10}^{-\text{5}}\) \(\text{N}\) to the right.

Three point charges form a right-angled triangle.Their charges are \(Q_1 = \text{4} \times \text{10}^{-\text{9}}\text{ C}=\text{4}\text{ nC}\), \(Q_2 = \text{6} \times \text{10}^{-\text{9}}\text{ C}=\text{6}\text{ nC}\)and \(Q_3 = -\text{3} \times \text{10}^{-\text{9}}\text{ C}=-\text{3}\text{ nC}\). The distance between\({Q}_{1}\) and \({Q}_{2}\)is \(\text{5} \times \text{10}^{-\text{2}}\) \(\text{m}\)and the distance between\({Q}_{1}\) and \({Q}_{3}\)is \(\text{3} \times \text{10}^{-\text{2}}\) \(\text{m}\). What is the net electrostatic force on \({Q}_{1}\) due to the other two charges if they are arranged as shown?

Step 1: Determine what is required

We need to calculate the net force on \({Q}_{1}\). This force is the sum of the two electrostatic forces - the forces of \({Q}_{2}\) on \({Q}_{1}\) and \({Q}_{3}\) on \({Q}_{1}\).

Step 2: Determine how to approach the problem

• We need to calculate, using Coulomb's law, the electrostatic force exerted on \(Q_1\) by \(Q_2\), and the electrostatic force exerted on \(Q_1\) by \(Q_3\).

• We then need to add up the two forces using our rules for adding vector quantities, because force is a vector quantity.

Step 3: Determine what is given

We are given all the charges and two of the distances.

Step 4: Calculate the magnitude of the forces.

The magnitude of the force exerted by \({Q}_{2}\) on \({Q}_{1}\), which we will call \(F_2\), is:

\begin{align*} F_2 & = k\frac{Q_1Q_2}{r^2} \\ &= (\text{9.0} \times \text{10}^{\text{9}})\frac{(\text{4} \times \text{10}^{-\text{9}})(\text{6} \times \text{10}^{-\text{9}})}{(\text{5} \times \text{10}^{-\text{2}})^2} \\ &= (\text{9.0} \times \text{10}^{\text{9}})\frac{(\text{4} \times \text{10}^{-\text{9}})(\text{6} \times \text{10}^{-\text{9}})}{(\text{25} \times \text{10}^{-\text{4}})} \\ &= \text{8.630} \times \text{10}^{-\text{5}}\text{ N} \end{align*}

The magnitude of the force exerted by \({Q}_{3}\) on \({Q}_{1}\), which we will call \(F_3\), is:

\begin{align*} F_3 & = k\frac{Q_1Q_3}{r^2} \\ &= (\text{9.0} \times \text{10}^{\text{9}})\frac{(\text{4} \times \text{10}^{-\text{9}})(\text{3} \times \text{10}^{-\text{9}})}{(\text{3} \times \text{10}^{-\text{2}})^2} \\ &= (\text{9.0} \times \text{10}^{\text{9}})\frac{(\text{4} \times \text{10}^{-\text{9}})(\text{3} \times \text{10}^{-\text{9}})}{(\text{9} \times \text{10}^{-\text{4}})} \\ &= \text{1.199} \times \text{10}^{-\text{4}}\text{ N} \end{align*}

Step 5: Vector addition of forces

This is a two-dimensional problem involving vectors. We have already solved many two-dimensional force problems and will use precisely the same procedure as before. Determine the vectors on the Cartesian plane, break them into components in the \(x\)- and \(y\)-directions, and then sum components in each direction to get the components of the resultant.

We choose the positive directions to be to the right (the positive \(x\)-direction) and up (the positive \(y\)-direction). We know the magnitudes of the forces but we need to use the signs of the charges to determine whether the forces are repulsive or attractive. Then we can use a diagram to determine the directions.

The force between \({Q}_{1}\) and \({Q}_{2}\) is repulsive (like charges). This means that it pushes \({Q}_{1}\) to the left, or in the negative \(x\)-direction.

The force between \({Q}_{1}\) and \({Q}_{3}\) is attractive(unlike charges) and pulls \({Q}_{1}\) in the positive \(y\)-direction.

We can redraw the diagram as a free-body diagram illustrating the forces to make sure we can visualise the situation:

Step 6: Resultant force

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^{2}_R & = F_{2}^{2} + F_{3}^{2} \text{by Pythagoras' theorem}\\ F_R &= \sqrt{(\text{8.630} \times \text{10}^{-\text{5}})^{2} + (\text{1.199} \times \text{10}^{-\text{4}})^{2}}\\ F_R &= \text{1.48} \times \text{10}^{-\text{4}}\text{ N} \end{align*} and the angle, \(\theta_R\) made with the \(x\)-axis can be found using trigonometry.

\begin{align*} \tan(\theta_R) &= \frac{\text{y-component}}{\text{x-component}} \\ \tan(\theta_R) &= \frac{\text{1.199} \times \text{10}^{-\text{4}}}{\text{8.630} \times \text{10}^{-\text{5}}} \\ \theta_R &= \tan^{-1}(\frac{\text{1.199} \times \text{10}^{-\text{4}}}{\text{8.630} \times \text{10}^{-\text{5}}}) \\ \theta_R &= \text{54.25}\text{°} {\text{ to 2 decimal places}} \end{align*}

The final resultant force acting on \(Q_1\) is \(\text{1.48} \times \text{10}^{-\text{4}}\) \(\text{N}\) acting at an angle of \(\text{54.25}\)\(\text{°}\) to the negative \(x\)-axis or \(\text{125.75}\)\(\text{°}\)to the positive \(x\)-axis.