IB Physics SL D.2.2 Coulomb's Law

Practise calculating electrostatic force between point charges, using centre separation and charge signs to establish magnitude and direction, then applying inverse-square scaling.

Exam points

calculate F = k|q₁q₂|/r² with charges in coulombs and centre-to-centre distance in metres
use charge signs separately to assign attraction or repulsion after finding the magnitude
compare configurations through direct charge scaling and inverse-square distance scaling

[Maximum number: 2]

The diagram shows two isolated electrons, X and Y , initially at rest in a vacuum. The initial separation of the electrons is 5.0 mm . The electrons subsequently move apart in the directions shown.

Figure for Question D.2.2—Coulomb’s law question 1 — IB Physics SL

Show that the initial electric force acting on each electron due to the other electron is approximately $9 \times 10^{-24} \mathrm{~N}$ .

(c) (i) use of $F=\frac{\left(1.60 \times 10^{-19}\right)^{2}}{4 \pi \varepsilon_{0}\left(5.0 \times 10^{-3}\right)^{2}}$ or $F=\frac{\left(1.60 \times 10^{-19}\right)^{2}}{\left(5.0 \times 10^{-3}\right)^{2}} \times 8.99 \times 10^{9}$ ;
$9.2 \times 10^{-24} \mathrm{~N} ;$

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D.2.2—Coulomb’s law question 1