To tackle this problem, we need to analyze how the presence of a uniform electric field affects the motion of charged particles compared to neutral particles. Let's break it down step by step.
Understanding the Motion of Charged and Neutral Particles
When a neutral particle is thrown at an angle, it follows a parabolic trajectory due to the influence of gravity alone. The horizontal range \( L \) is determined by its initial velocity and the angle of projection. However, when we introduce a charged particle into the scenario, the dynamics change because the charged particle experiences a force due to the electric field.
Force on Charged Particles
The force \( F \) acting on a charged particle in an electric field \( E \) is given by:
F = qE
This force will affect the vertical motion of the charged particle, causing it to deviate from the parabolic path of the neutral particle. The electric force acts in the direction of the electric field, which we can assume is directed upwards for this explanation.
Analyzing the Range of the Charged Particles
For the neutral particle, the range \( R \) is given by the formula:
R = \frac{v^2 \sin(2\theta)}{g}
where \( v \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.
Now, for the charged particle with charge \( q \), the electric force modifies its vertical motion. The effective acceleration in the vertical direction becomes \( g - \frac{qE}{m} \). This results in a shorter range because the particle is effectively "pulled" upwards by the electric field, reducing the time it spends in the air. Given that this particle has a range of \( L/2 \), we can conclude that the electric field has a significant impact on its trajectory.
Considering the Charged Particle with Charge 2q
Now, let's analyze the particle with charge \( 2q \). The force acting on this particle will be:
F = 2qE
This means that the vertical acceleration will be affected even more significantly, as it will now be \( g - \frac{2qE}{m} \). The increased electric force will cause the particle to be pulled upwards more strongly, further reducing its time of flight compared to the particle with charge \( q \).
Calculating the New Range
Since the range of the particle with charge \( q \) is \( L/2 \), we can infer that the range of the particle with charge \( 2q \) will be reduced even further. The relationship between the charge and the range can be thought of as inversely proportional, given that the electric force increases with charge.
Thus, if the range of the first charged particle is \( L/2 \), the range of the second charged particle with double the charge will be:
Range of particle with charge 2q = \frac{L/2}{2} = \frac{L}{4}
Final Answer
Therefore, the range of the particle with charge \( 2q \) is \( L/4 \). The correct option is (d) \( L/4 \).
