A uniform electric field $$\vec{E}$$ exists between the plates of a charged condenser. A charged particle enters the space between the plates and perpendicular to $$\vec{E}$$. The path of the particle between the plates is a:

A uniform electric field $$\vec{E}$$ exists between the plates of a charged condenser. A charged particle enters this space perpendicular to $$\vec{E}$$. To determine the path, we analyze the forces and motion.

The force on a charged particle in an electric field is given by $$\vec{F} = q \vec{E}$$, where $$q$$ is the charge of the particle. Since the electric field is uniform, the force is constant. Assume the electric field $$\vec{E}$$ is directed along the y-axis, so $$\vec{E} = E \hat{j}$$. The particle enters perpendicular to $$\vec{E}$$, meaning its initial velocity $$\vec{u}$$ is along the x-axis, so $$\vec{u} = u \hat{i}$$.

The force $$\vec{F} = q E \hat{j}$$ causes an acceleration. By Newton's second law, $$\vec{F} = m \vec{a}$$, so:

$$ m \vec{a} = q E \hat{j} $$

Thus, the acceleration $$\vec{a}$$ is:

$$ \vec{a} = \frac{q E}{m} \hat{j} $$

This acceleration is constant and acts only in the y-direction. There is no acceleration in the x-direction because no force acts there. So, the components of acceleration are:

$$ a_x = 0 $$

$$ a_y = \frac{q E}{m} $$

Assume the particle starts at the origin at time $$t = 0$$, with initial position $$x(0) = 0$$, $$y(0) = 0$$, and initial velocity components $$u_x = u$$, $$u_y = 0$$.

Using the equations of motion:

For the x-direction (constant velocity since $$a_x = 0$$):

$$ v_x = u_x + a_x t = u + 0 = u $$

$$ x = u_x t + \frac{1}{2} a_x t^2 = u t + 0 = u t $$

For the y-direction (constant acceleration $$a_y = \frac{q E}{m}$$):

$$ v_y = u_y + a_y t = 0 + \frac{q E}{m} t = \frac{q E}{m} t $$

$$ y = u_y t + \frac{1}{2} a_y t^2 = 0 + \frac{1}{2} \left( \frac{q E}{m} \right) t^2 = \frac{1}{2} \frac{q E}{m} t^2 $$

We have parametric equations:

$$ x = u t \quad \text{(1)} $$

$$ y = \frac{1}{2} \frac{q E}{m} t^2 \quad \text{(2)} $$

To find the trajectory, eliminate the parameter $$t$$. From equation (1), solve for $$t$$:

$$ t = \frac{x}{u} $$

Substitute this into equation (2):

$$ y = \frac{1}{2} \frac{q E}{m} \left( \frac{x}{u} \right)^2 = \frac{1}{2} \frac{q E}{m} \frac{x^2}{u^2} $$

Simplify:

$$ y = \left( \frac{q E}{2 m u^2} \right) x^2 $$

This equation $$y = k x^2$$, where $$k = \frac{q E}{2 m u^2}$$ is a constant, represents a parabola. Therefore, the path of the particle is parabolic.

Comparing with the options:

A. straight line

B. hyperbola

C. parabola

D. circle

Hence, the correct answer is Option C.