1) Two alpha particles are fired directly at each other in a vacuum. Both are originally travelling with a speed of 2.4times 10^5m/s 9 marks a).Find their separation at closest approach . (5) [2.4times 10^-12m] b) What is the electrostatic force of repulsion between the two charged particles at the position of closest approach? (2) [1.6times 10^-4N] c) Sketch what the electric field would look like in the space around and between the two charges at the closest separation . (2)

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1) Two alpha particles are fired directly at each other in a vacuum. Both are originally travelling with a speed of 2.4times 10^5m/s 9 marks a).Find their separation at closest approach . (5) [2.4times 10^-12m] b) What is the electrostatic force of repulsion between the two charged particles at the position of closest approach? (2) [1.6times 10^-4N] c) Sketch what the electric field would look like in the space around and between the two charges at the closest separation . (2)

Answer 1

**1a) Finding the separation at closest approach:*** **Concept:** At the closest approach, all the initial kinetic energy of the alpha particles is converted into electrostatic potential energy. This is because the particles momentarily come to rest before repelling each other.* **Formulae:** * Kinetic Energy (KE) = (1/2)mv² * Electrostatic Potential Energy (PE) = kq₁q₂/r * where: * m = mass of an alpha particle (6.64 x 10⁻²⁷ kg) * v = initial speed of each alpha particle (2.4 x 10⁵ m/s) * k = Coulomb's constant (8.99 x 10⁹ Nm²/C²) * q₁ = q₂ = charge of an alpha particle (2 * elementary charge = 2 * 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C) * r = separation at closest approach (what we want to find)* **Calculations:** 1. KE of one alpha particle = (1/2) * (6.64 x 10⁻²⁷ kg) * (2.4 x 10⁵ m/s)² = 1.91 x 10⁻¹⁶ J 2. Total KE of both alpha particles = 2 * 1.91 x 10⁻¹⁶ J = 3.82 x 10⁻¹⁶ J 3. At closest approach, Total KE = PE, so: 3.82 x 10⁻¹⁶ J = (8.99 x 10⁹ Nm²/C²) * (3.2 x 10⁻¹⁹ C)² / r 4. Solving for r: r = (8.99 x 10⁹ Nm²/C²) * (3.2 x 10⁻¹⁹ C)² / (3.82 x 10⁻¹⁶ J) ≈ 2.4 x 10⁻¹² m**1b) Electrostatic force at closest approach:*** **Concept:** Use Coulomb's Law to calculate the force.* **Formula:** F = kq₁q₂/r²* **Calculations:** F = (8.99 x 10⁹ Nm²/C²) * (3.2 x 10⁻¹⁹ C)² / (2.4 x 10⁻¹² m)² ≈ 1.6 x 10⁻⁴ N**1c) Sketch of the electric field:*** **Concept:** The electric field lines originate from positive charges and terminate on negative charges (or extend to infinity if there are no negative charges). The field lines should be denser where the field is stronger.* **Sketch:** Imagine the two alpha particles (positive charges) close together. * Draw field lines radiating outwards from each alpha particle. * The lines should curve away from each other between the two particles, indicating repulsion. There should be no field lines directly between the particles connecting them. * The field lines should be most dense close to the charges, indicating a stronger field there. Further away, the lines should become less dense. A simple representation would look something like this, keeping in mind this is a 2D representation of a 3D field: ``` + + / \ / \ / \ / \ \ / \ / \ / \ / + + ``` The '+' symbols represent the alpha particles. The lines represent the electric field lines. The field is strongest near the charges and weakens as you move further away. The field lines repel each other in the space between the charges.

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