Worked examples · Sheet 1

Forces between two point charges

Theory used: Coulomb's law Point charge Quantization of charge Dependence on distance Action and reaction between two charges

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Two point charges $q_{1} = + 3.0 μ C$ and $q_{2} = - 5.0 μ C$ are in vacuum at distance $r = 0.30 m$ . Compute the magnitude of the force that each charge exerts on the other and determine its direction and orientation.

F = k \frac{∣ q _{1} ∣ ∣ q _{2} ∣}{r ^{2}}

F = 8.99 \cdot 1 0^{9} \cdot \frac{3.0 \cdot 1 0 ^{- 6} \cdot 5.0 \cdot 1 0 ^{- 6}}{0.3 0 ^{2}} N = \frac{0.135}{0.090} N \approx 1.5 N

The two charges are point charges at rest in vacuum, so Coulomb’s law applies. The magnitude of the force is computed with the absolute values of the charges, and the orientation is obtained from the sign rule. The charges have opposite signs, therefore they attract each other, and the force on $q_{2}$ is directed along the joining line toward $q_{1}$ , while the force on $q_{1}$ is directed toward $q_{2}$ .

By action and reaction between two charges the two forces have the same magnitude, $1.5 N$ , even though the charges have different absolute values. The result has two significant figures, like the data.

Exercise 2

Two identical small spheres, with the same positive charge $q$ , are in vacuum at distance $r = 5.0 cm$ and repel each other with a force of magnitude $F = 0.90 N$ . The spheres are treated as point charges. Compute $q$ , the number of electrons missing from each sphere and the magnitude of the force when the distance triples.

F = k \frac{q ^{2}}{r ^{2}} ⟹ q = r \frac{F}{k}, n = \frac{q}{e}, F^{'} = k \frac{q ^{2}}{( 3 r ) ^{2}} = \frac{F}{9}

q n F^{'} = 0.050 \cdot \frac{0.90}{8.99 \cdot 1 0 ^{9}} C = 0.050 \cdot 1.0 \cdot 1 0^{- 5} C = 5.0 \cdot 1 0^{- 7} C = 0.50 μ C = \frac{5.0 \cdot 1 0 ^{- 7}}{1.60 \cdot 1 0 ^{- 19}} \approx 3.1 \cdot 1 0^{12} = \frac{0.90}{9} N = 0.10 N

The spheres have the same charge, so in Coulomb’s law the product of the absolute values of the charges is $q^{2}$ . We obtain $q$ by dividing by $k$ , multiplying by $r^{2}$ and taking the positive root, because the charge is positive by hypothesis. The distance must be expressed in meters, $5.0 cm = 0.050 m$ , since $k$ is expressed in International System units.

A positively charged sphere has lost electrons; by the quantization of charge the number of missing electrons is the ratio between the charge and the elementary charge. Finally the force is inversely proportional to the square of the distance, therefore when the distance triples the force becomes one ninth, without any need to recompute the charge.

Exercise 3

In vacuum the charge $q_{1} = + 2.0 μ C$ is at the origin and the charge $q_{2} = - 3.0 μ C$ at the point $(0.40 m; 0.30 m)$ . Compute in components the force $F_{12}$ that $q_{1}$ exerts on $q_{2}$ , its magnitude and the force $F_{21}$ that $q_{2}$ exerts on $q_{1}$ .

The force that

q_{1}

exerts on

q_{2}

is directed along the joining line toward

q_{1}

, because the charges have opposite signs

F_{12} = k \frac{q _{1} q _{2}}{∣ r _{2} - r _{1} ∣ ^{3}} (r_{2} - r_{1}), F_{21} = - F_{12}

r_{2} - r_{1} k \frac{q _{1} q _{2}}{∣ r _{2} - r _{1} ∣ ^{3}} F_{12} F_{12} F_{21} = (0.40 \overset{}{^} + 0.30 \overset{}{^}) m, ∣ r_{2} - r_{1} ∣ = 0.4 0^{2} + 0.3 0^{2} m = 0.50 m = 8.99 \cdot 1 0^{9} \cdot \frac{2.0 \cdot 1 0 ^{- 6} \cdot ( - 3.0 \cdot 1 0 ^{- 6} )}{0.5 0 ^{3}} N/m \approx - 0.432 N/m \approx - 0.432 \cdot (0.40 \overset{}{^} + 0.30 \overset{}{^}) N \approx (- 0.17 \overset{}{^} - 0.13 \overset{}{^}) N = 0.432 \cdot 0.50 N \approx 0.22 N \approx (0.17 \overset{}{^} + 0.13 \overset{}{^}) N

The positions are given in coordinates, so we use the vector form of Coulomb’s law, which gives the components with the correct sign. The vector $r_{2} - r_{1}$ goes from $q_{1}$ to $q_{2}$ and has length $0.50 m$ , because the legs measure $0.40 m$ and $0.30 m$ . The coefficient $k \frac{q _{1} q _{2}}{r ^{3}}$ is negative, since the charges have opposite signs; therefore $F_{12}$ has orientation opposite to $r_{2} - r_{1}$ , that is it points from $q_{2}$ toward $q_{1}$ , as an attraction requires.

The magnitude is the absolute value of the coefficient times the distance, and it coincides with $k \frac{∣ q _{1} ∣ ∣ q _{2} ∣}{r ^{2}} = 8.99 \cdot 1 0^{9} \cdot \frac{6.0 \cdot 1 0 ^{- 12}}{0.25} N \approx 0.22 N$ , which is a check on the calculation. By action and reaction between two charges the force on $q_{1}$ is the opposite of $F_{12}$ and points toward $q_{2}$ .