Nearest approach.

Classical Mechanics Level 3

A small particle of mass m m and charge q q whose dimensions are negligible as compared to the radius of the ring approaches the ring from very far (assumed to be from infinity) with initial velocity v 0 v_0 along the axis of the ring as shown. The ring has a radius R R and also has mass m m and charge q q uniformly distributed over it. Assuming gravitational forces to be absent and the system is void of any other forces except the electrostatic forces acting between the bodies, find the closest distance of approach between the particle and the plane of the ring using the data given below.

m m = 2 kg 2 \text{ kg}
q q = 20 μ C 20 \ \mu \text{C}
v 0 v_0 = 1 ms − 1 1 \text{ ms}^{-1}
R R = 6 m 6 \text{ m}
1 4 π ϵ 0 \dfrac{1}{4\pi \epsilon_0} = 9 × 1 0 9 Nm 2 C − 2 9 \times 10^{9} \text{ Nm}^2 \text{C}^{-2}

Enter your answer rounded to the closest integer value.


The answer is 4.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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2 solutions

Steven Chase
Mar 31, 2018

The potential energy of the system when the two objects are separated by a distance D D is:

U = k q 2 R 2 + D 2 \large{U = \frac{k q^2}{\sqrt{R^2 + D^2}}}

Note that the two objects have the same mass, and experience the same forces in opposite directions. Therefore, they will experience the same changes in velocity (also in opposite directions). The kinetic energy at any particular instant is:

E = 1 2 m ( v 0 − α ) 2 + 1 2 m α 2 \large{E = \frac{1}{2} m (v_0 - \alpha)^2 + \frac{1}{2} m \alpha^2}

The closest approach occurs when the kinetic energy is minimized, meaning that the greatest amount of the initial kinetic energy has been converted into potential energy. Differentiate the kinetic energy expression to find the corresponding value of α \alpha .

E = − m ( v 0 − α ) + m α = 0 ⟹ α = v 0 2 \large{E = -m (v_0 - \alpha) + m \alpha = 0 \\ \implies \alpha = \frac{v_0}{2}}

It is therefore clear that the closest approach corresponds to half the initial kinetic energy being stored as potential energy:

1 4 m v 0 2 = k q 2 R 2 + D C 2 \large{\frac{1}{4} m v_0^2 = \frac{k q^2}{\sqrt{R^2 + D_C^2}}}

Re-arranging gives:

D C = 16 k 2 q 4 m 2 v 0 4 − R 2 ≈ 3.98 \large{D_C = \sqrt{\frac{16 k^2 q^4}{m^2 v_0^4} - R^2} \approx 3.98}

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Ashutosh Sharma
Mar 31, 2018

conserve momentum first then energy

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