Question 878: Electrostats & Capacitance

Question

Two particles $$A$$ and $$B$$ having same mass have charge $$+q$$ and $$+4 q$$ respectively. When they are allowed to fall from rest through same electric potential difference, the ratio of their speeds '$$V_A$$' to '$$\mathrm{V}_{\mathrm{B}}$$' will become

Accepted Answer

To find the ratio of the speeds $$V_A$$ to $$\mathrm{V}_{\mathrm{B}}$$ of two particles $$A$$ and $$B$$ after falling through the same electric potential difference, we should consider the work done by the electric field on each particle due to the potential difference. This work done will convert into kinetic energy for each particle assuming they start from rest.
The work done on a charge by an electric potential difference $$V$$ is given by the formula:
$$
W = qV
$$
where
$$W$$ is the work done or energy gained,
$$q$$ is the charge of the particle, and
$$V$$ is the electric potential difference.
Since the initial kinetic energy of both particles is zero (as they are initially at rest), the entire work done on each will convert into its kinetic energy. The kinetic energy ($$K$$) of a particle is given by:
$$
K = \frac{1}{2}mv^2
$$
where
$$m$$ is the mass of the particle, and
$$v$$ is its speed.
Thus, for particle $$A$$ with charge $$+q$$,
$$
qV = \frac{1}{2}mV_A^2
$$
And for particle $$B$$ with charge $$+4q$$,
$$
4qV = \frac{1}{2}mV_B^2
$$
Now, to find the ratio of their speeds, $$V_A$$ to $$V_B$$, we can set up a ratio using the above equations. Dividing the equation for particle $$A$$ by the equation for particle $$B$$,
$$
\frac{qV}{4qV} = \frac{\frac{1}{2}mV_A^2}{\frac{1}{2}mV_B^2}
$$
Simplifying the equation gives,
$$
\frac{1}{4} = \frac{V_A^2}{V_B^2}
$$
Taking the square root of both sides,
$$
\frac{V_A}{V_B} = \frac{1}{2}
$$
This means the ratio of their speeds $$V_A : V_B$$ is $$1:2$$.
Therefore, the correct answer is:
Option A: $$1: 2$$.

Answer

$$1: 2$$

Answer

$$2: 1$$

Answer

$$1: 4$$

Answer

$$4: 1$$

Electrostats & Capacitance

Two particles $A and B having same mass have charge +q and +4 q respectively. When they are allowed to fall from rest through same electric potential difference, the ratio of their speeds 'V_A' to '\mathrm{V}_{\mathrm{B}}$' will become

Choose the correct answer: