Electric Charge (DP IB Physics: HL): Revision Note

Remember the saying: “Opposites attract”.

Four identical metal spheres have charges of q_A = −8.0 µC, q_B = −2.0 µC, q_C = +5.0 µC, and q_D = +12.0 µC.

(a) Two of the spheres are brought into contact briefly, and then they are separated. Which spheres are they if the final charge on each one is +5.0 µC?

(b) All four spheres are brought into contact briefly and then separated. What is the final charge on each sphere?

(c) How many electrons would have to be added to one of the spheres in (b) to make it electrically neutral? 

Answer:

(a)

Step 1: Apply the principle of conservation of charge to the scenario

• When two charged spheres come into contact, the charges are shared between them until they are evenly distributed i.e. both spheres have the same charge

• The charge on each sphere is equal to the average of the two charges

$Q_{final}=\frac{Q_1+Q_2}{2}$

Step 2: Determine the charge on each sphere

• For the average charge to be +5 μC, the sum of the two charges must be +10 μC

• This can only be achieved with charges q_B = −2.0 µC and q_D = +12.0 µC

$Q_{final}=\frac{12.0-2.0}{2}=+5.0$ μC

(b)

Step 1: Apply the principle of conservation of charge to the scenario

• The charge on each sphere is equal to the average of the four charges (i.e. the total charge is equally distributed between all four spheres)

$Q_{final}=\frac{Q_1+Q_2+Q_3+Q_4}{4}$

Step 2: Determine the charge on each sphere

• The average charge on each sphere is

$Q_{final}=\frac{12.0+5.0-2.0-8.0}{4}=+1.75$ μC

Note: you would also get the same result if you used q_B = q_D = +5.0 µC

(c)

Step 1: Recall the charge of an electron and that charge is quantised

• Electrons have a charge of e = −1.60 × 10⁻¹⁹ C

• Therefore, the number of electrons required is

$\mathrm{number}\mathrm{of}\mathrm{electrons}=\frac{\mathrm{charge}\mathrm{on}\mathrm{sphere}}{e}$

Step 2: Determine the number of electrons required

$\mathrm{number}\mathrm{of}\mathrm{electrons}=\frac{1.75\times {10}^{-6}}{1.60\times {10}^{-19}}=1.094\times {10}^{13}$

• Therefore, 1.1 × 10¹³ electrons are required to neutralise one of the charges

The law of conservation of charge is also important when considering systems of particles and processes, such as nuclear decay

For example, in beta decay, when a neutron decays into a proton, an electron must also be produced to balance the charges

$\mathrm{n}\to {\mathrm{p}}^{+}+{\mathrm{e}}^{-}+{\overline{\mathrm{\nu }}}_{\mathrm{e}}$

charge on LHS = 0

charge on RHS = 1 + (−1) + 0 = 0