Worked examples · Sheet 2

Charges after contact between conducting spheres

Theory used: Conservation of charge Quantization of charge Charging by contact Contact between two identical conducting spheres Coulomb's law

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Two identical conducting spheres $A$ and $B$ have charges $q_{A} = + 8.0 nC$ and $q_{B} = - 2.0 nC$ . The spheres are brought into contact and then separated. Compute the final charge of each sphere, determine in which direction the electrons move and how many there are.

q_{A}^{'} = q_{B}^{'} = \frac{q _{A} + q _{B}}{2}, Δ q_{B} = q_{B}^{'} - q_{B}, n = \frac{∣Δ q _{B} ∣}{e}

q_{A}^{'} = q_{B}^{'} Δ q_{B} n = \frac{8.0 - 2.0}{2} nC = + 3.0 nC = (3.0 - (- 2.0)) nC = + 5.0 nC = \frac{5.0 \cdot 1 0 ^{- 9}}{1.60 \cdot 1 0 ^{- 19}} \approx 3.1 \cdot 1 0^{10}

The spheres are identical conductors, so after the contact by symmetry they have the same charge, and by conservation of charge the sum of the final charges equals the initial sum $+ 6.0 nC$ . Each sphere therefore ends up with $+ 3.0 nC$ .

In metallic conductors only the electrons move. The charge of $B$ increases by $5.0 nC$ , and an increasing charge corresponds to electrons leaving; hence the electrons move from sphere $B$ to sphere $A$ . The charge of $A$ decreases by the same amount, as it must. The number of electrons is obtained from the quantization of charge, by dividing the transferred charge by $e$ .

Exercise 2

Two identical small conducting spheres, with charges $q_{A} = + 5.0 μ C$ and $q_{B} = - 1.0 μ C$ , are in vacuum at distance $r = 0.30 m$ . The spheres are brought into contact and then returned to the same distance. Compute the force between the spheres before and after the contact, treating them as point charges.

F = k \frac{∣ q _{A} ∣ ∣ q _{B} ∣}{r ^{2}}, q^{'} = \frac{q _{A} + q _{B}}{2}, F^{'} = k \frac{q ^{'2}}{r ^{2}}

F q^{'} F^{'} = 8.99 \cdot 1 0^{9} \cdot \frac{5.0 \cdot 1 0 ^{- 6} \cdot 1.0 \cdot 1 0 ^{- 6}}{0.3 0 ^{2}} N \approx 0.50 N = \frac{5.0 - 1.0}{2} μ C = + 2.0 μ C = 8.99 \cdot 1 0^{9} \cdot \frac{( 2.0 \cdot 1 0 ^{- 6} ) ^{2}}{0.3 0 ^{2}} N \approx 0.40 N

Before the contact the charges have opposite signs, so the force is attractive and its magnitude is computed with Coulomb’s law. During the contact the total charge $+ 4.0 μ C$ is conserved and divides into equal parts between the two identical spheres. After the contact the spheres have the same positive charge, therefore the force becomes repulsive.

The magnitude goes from $0.50 N$ to $0.40 N$ . The ratio between the two magnitudes is the ratio between the products of the charges, $\frac{4.0}{5.0} = 0.80$ , because the distance is the same; this comparison checks the two calculations.

Exercise 3

Three identical conducting spheres $A$ , $B$ and $C$ are isolated. Initially $A$ has charge $Q = + 16 nC$ , and $B$ and $C$ are neutral. $A$ is brought into contact with $B$ , then $A$ with $C$ , finally $B$ with $C$ , separating the spheres after each contact. Compute the final charges and the force between $B$ and $C$ when their centers are $3.0 cm$ apart, treating them as point charges.

A with B : A with C : B with C : q_{A} = q_{B} = \frac{16 + 0}{2} nC = 8.0 nC q_{A} = q_{C} = \frac{8.0 + 0}{2} nC = 4.0 nC q_{B} = q_{C} = \frac{8.0 + 4.0}{2} nC = 6.0 nC

F = k \frac{q _{B} q _{C}}{r ^{2}} = 8.99 \cdot 1 0^{9} \cdot \frac{( 6.0 \cdot 1 0 ^{- 9} ) ^{2}}{( 3.0 \cdot 1 0 ^{- 2} ) ^{2}} N \approx 3.6 \cdot 1 0^{- 4} N

At each contact between two identical spheres the rule for contact between two identical conducting spheres applies, according to which each sphere takes the average of the two charges that took part in the contact. The sphere that does not take part keeps its own charge. The final charges are $q_{A} = 4.0 nC$ and $q_{B} = q_{C} = 6.0 nC$ , with sum $16 nC$ , equal to the initial charge, as conservation of charge requires.

The spheres $B$ and $C$ have charges of the same sign, so they repel each other, and the magnitude of the force is given by Coulomb’s law with the distance in meters.

Exercise 4

Two identical small conducting spheres, placed in vacuum at distance $r = 0.50 m$ , attract each other with a force of magnitude $F_{1} = 0.432 N$ . The spheres are brought into contact and then returned to the same distance, and now they repel each other with a force of magnitude $F_{2} = 0.144 N$ . Compute the initial charges, treating the spheres as point charges.

⎩ ⎨ ⎧ q_{1} q_{2} = - \frac{F _{1} r ^{2}}{k} (\frac{q _{1} + q _{2}}{2})^{2} = \frac{F _{2} r ^{2}}{k} ⟹ {q_{1} q_{2} = p q_{1} + q_{2} = s with p = - \frac{F _{1} r ^{2}}{k}, s = \pm 2 r \frac{F _{2}}{k}

p s = - \frac{0.432 \cdot 0.5 0 ^{2}}{8.99 \cdot 1 0 ^{9}} C^{2} \approx - 1.20 \cdot 1 0^{- 11} C^{2} = \pm 2 \cdot 0.50 \cdot \frac{0.144}{8.99 \cdot 1 0 ^{9}} C \approx \pm 4.00 \cdot 1 0^{- 6} C

t^{2} - s t + p = 0 ⟹ t = \frac{4.00 \pm 4.0 0 ^{2} + 4 \cdot 12.0}{2} μ C = \frac{4.00 \pm 8.00}{2} μ C

q_{1} = + 6.0 μ C, q_{2} = - 2.0 μ C or q_{1} = - 6.0 μ C, q_{2} = + 2.0 μ C

Before the contact the force is attractive, so the charges have opposite signs and their product is negative; Coulomb’s law then gives $q_{1} q_{2} = - \frac{F _{1} r ^{2}}{k}$ . After the contact each sphere has the charge $\frac{q _{1} + q _{2}}{2}$ , by contact between two identical conducting spheres, and Coulomb’s law applied to the repulsive force gives the square of that charge. The two equations must hold together and form a system in the unknowns $q_{1}$ and $q_{2}$ , of which the product and the sum are known, up to the sign of the sum.

Two numbers with sum $s$ and product $p$ are the solutions of the equation $t^{2} - s t + p = 0$ . With $s = + 4.00 μ C$ and $p = - 12.0 (μ C)^{2}$ the solutions are $+ 6.0 μ C$ and $- 2.0 μ C$ ; with $s = - 4.00 μ C$ the opposite charges are obtained. The forces depend only on the products of the charges, therefore the force measurements do not distinguish the two solutions; hence the charges are, in absolute value, $6.0 μ C$ and $2.0 μ C$ , with opposite signs.

As a check, with $+ 6.0 μ C$ and $- 2.0 μ C$ the initial force is $8.99 \cdot 1 0^{9} \cdot \frac{12 \cdot 1 0 ^{- 12}}{0.25} N \approx 0.432 N$ , and after the contact each charge is $+ 2.0 μ C$ and the force is $8.99 \cdot 1 0^{9} \cdot \frac{4.0 \cdot 1 0 ^{- 12}}{0.25} N \approx 0.144 N$ .