Worked examples · Sheet 3

Forces on collinear charges

Theory used: Superposition principle Charges on a line Coulomb's law Sum of the internal forces

Exercise 1

On the $x$ -axis, in vacuum, there are the point charges $q_{1} = + 2.0 μ C$ at $x_{1} = 0$ and $q_{2} = - 4.0 μ C$ at $x_{2} = 0.30 m$ . Compute the total force on the charge $q_{0} = + 1.0 μ C$ placed at $x_{0} = 0.10 m$ .

F_{x} = k \frac{q _{1} q _{0}}{( x _{0} - x _{1} ) ^{2}} + k \frac{∣ q _{2} ∣ q _{0}}{( x _{2} - x _{0} ) ^{2}}

F_{1} F_{2} F_{x} = 8.99 \cdot 1 0^{9} \cdot \frac{2.0 \cdot 1 0 ^{- 6} \cdot 1.0 \cdot 1 0 ^{- 6}}{0.1 0 ^{2}} N \approx 1.80 N = 8.99 \cdot 1 0^{9} \cdot \frac{4.0 \cdot 1 0 ^{- 6} \cdot 1.0 \cdot 1 0 ^{- 6}}{0.2 0 ^{2}} N \approx 0.90 N = F_{1} + F_{2} \approx 2.7 N

By the superposition principle the total force on $q_{0}$ is the sum of the forces that $q_{1}$ and $q_{2}$ would exert alone, and all the forces are directed along the $x$ -axis. The charge $q_{1}$ is positive like $q_{0}$ and lies to its left, so it repels it to the right. The charge $q_{2}$ is negative and lies to the right of $q_{0}$ , so it attracts it to the right. The two contributions have the same orientation and add up.

The total force has magnitude $2.7 N$ and is directed to the right, that is toward $q_{2}$ . The contribution of $q_{2}$ is half that of $q_{1}$ , because $q_{2}$ has twice the absolute value and twice the distance from $q_{0}$ , so its contribution is multiplied by $\frac{2}{2 ^{2}} = \frac{1}{2}$ .

Exercise 2

On the $x$ -axis, in vacuum, there are the point charges $q_{1} = + 3.0 μ C$ at $x_{1} = 0$ and $q_{2} = - 3.0 μ C$ at $x_{2} = 0.20 m$ . Compute the total force on the charge $q_{0} = + 2.0 μ C$ placed at $x_{0} = 0.40 m$ .

F_{x} = k \frac{q _{1} q _{0}}{( x _{0} - x _{1} ) ^{2}} - k \frac{∣ q _{2} ∣ q _{0}}{( x _{0} - x _{2} ) ^{2}}

F_{1} F_{2} F_{x} = 8.99 \cdot 1 0^{9} \cdot \frac{3.0 \cdot 1 0 ^{- 6} \cdot 2.0 \cdot 1 0 ^{- 6}}{0.4 0 ^{2}} N \approx 0.337 N = 8.99 \cdot 1 0^{9} \cdot \frac{3.0 \cdot 1 0 ^{- 6} \cdot 2.0 \cdot 1 0 ^{- 6}}{0.2 0 ^{2}} N \approx 1.349 N = F_{1} - F_{2} \approx - 1.0 N

Now $q_{0}$ lies to the right of both charges. The charge $q_{1}$ , positive, repels it to the right; the charge $q_{2}$ , negative, attracts it to the left. By the rule for charges on a line the first contribution is positive and the second negative.

The two charges have the same absolute value, and $q_{2}$ is at half the distance of $q_{1}$ , so its contribution is four times larger. The total force has a negative component, that is it is directed to the left, with magnitude about $1.0 N$ .

Exercise 3

On the $x$ -axis, in vacuum, there are the point charges $q_{1} = + 1.0 μ C$ at $x_{1} = 0$ , $q_{2} = + 2.0 μ C$ at $x_{2} = 0.10 m$ and $q_{3} = - 3.0 μ C$ at $x_{3} = 0.30 m$ . Compute the total force on each charge and verify that their sum is zero.

F_{12} = k \frac{∣ q _{1} q _{2} ∣}{0.1 0 ^{2}}, F_{13} = k \frac{∣ q _{1} q _{3} ∣}{0.3 0 ^{2}}, F_{23} = k \frac{∣ q _{2} q _{3} ∣}{0.2 0 ^{2}}

F_{12} F_{13} F_{23} = 8.99 \cdot 1 0^{9} \cdot \frac{2.0 \cdot 1 0 ^{- 12}}{0.010} N \approx 1.798 N = 8.99 \cdot 1 0^{9} \cdot \frac{3.0 \cdot 1 0 ^{- 12}}{0.090} N \approx 0.300 N = 8.99 \cdot 1 0^{9} \cdot \frac{6.0 \cdot 1 0 ^{- 12}}{0.040} N \approx 1.349 N

F_{1, x} F_{2, x} F_{3, x} F_{1, x} + F_{2, x} + F_{3, x} = - F_{12} + F_{13} \approx (- 1.798 + 0.300) N \approx - 1.50 N = + F_{12} + F_{23} \approx (1.798 + 1.349) N \approx + 3.15 N = - F_{13} - F_{23} \approx (- 0.300 - 1.349) N \approx - 1.65 N \approx (- 1.50 + 3.15 - 1.65) N = 0

Total forces on the three charges, drawn to scale: the sum of the three forces is zero

The three charges form three pairs, and for each pair the magnitude of the force is computed only once with Coulomb’s law, because by action and reaction the two charges of the pair experience forces of equal magnitude and opposite orientations. The pair $q_{1}$ , $q_{2}$ is repulsive; the pairs $q_{1}$ , $q_{3}$ and $q_{2}$ , $q_{3}$ are attractive.

On $q_{1}$ the charge $q_{2}$ pushes to the left and $q_{3}$ pulls to the right; on $q_{2}$ the charge $q_{1}$ pushes to the right and $q_{3}$ also pulls to the right; on $q_{3}$ both positive charges pull to the left. The total forces are therefore $1.50 N$ to the left on $q_{1}$ , $3.15 N$ to the right on $q_{2}$ and $1.65 N$ to the left on $q_{3}$ .

The sum is zero, as predicted by the sum of the internal forces: each magnitude $F_{ij}$ appears twice, with opposite signs. The check confirms the calculations.