Question

A point charge of mass m and charge Q and another point charge of mass m but charge 2Q are released on a frictionless table. If the charge Q has an initial acceleration ${\mathbf{a}}_{\mathbf{0}}$, what will be the acceleration of 2Q: ${\mathbf{a}}_{\mathbf{0}}\mathbf{}\mathbf{,}\mathbf{2}{\mathbf{a}}_{\mathbf{0}}\mathbf{,}\mathbf{}{\mathbf{a}}_{\mathbf{0}}\mathbf{/}\mathbf{2}$or ${\mathbf{a}}_{\mathbf{0}}\mathbf{}\mathbf{/}\mathbf{4}$? Explain.

Short answer

The initial acceleration is${\mathrm{a}}_0$.

Step-by-step solution

Definition of the acceleration

The rate at which an object's velocity changes with function of time is called acceleration.

The acceleration when the velocity is zero is called initial acceleration.

Calculation of the force exerted on both of the particles

Given that the table is frictionless which gives the only force exerted on both of the particles whose mass "m" is the same, horizontally, is the electric force due to their charges.

The electric force is,

$$\begin{gathered} \mathrm{F}=\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{R}}^2} \\ =\frac{\mathrm{k}\left(\mathrm{Q}\right)\left(2\mathrm{Q}\right)}{{\mathrm{R}}^2} \\ =\frac{2{\mathrm{kQ}}^2}{{\mathrm{R}}^2} \end{gathered}$$

The electric force exerted on them is the same but in a different direction.

Explanation for finding the initial acceleration for both of the particles

From the Newton’s second law, the relation between force and the acceleration is$\mathrm{F}=\mathrm{ma}$, which implies that.

As both, the particles are affected by the same magnitude of force, and the mass, then their initial acceleration is also the same. Hence, the initial acceleration is$\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}$.