Question

An electron is moving with velocity \(v_0 \hat{i}\). If a uniform electric field \(\vec{E} = E_0 \hat{j}\) is set up in the region, the electron will:

• A. describe a circular path
• B. describe a helical path
• C. describe a parabolic path
• D. continue moving without any deviation

Hint:

Remember: Perpendicular Electric Field \(\rightarrow\) Parabolic path. Perpendicular Magnetic Field \(\rightarrow\) Circular path.

Answer 1

The Correct Option is C

The given problem involves an electron moving with an initial velocity \(v_0 \hat{i}\) in the presence of a uniform electric field \(\vec{E} = E_0 \hat{j}\). Let us analyze the motion of the electron step-by-step:

1. The electron initially moves with a velocity \(v_0 \hat{i}\), which means it is moving along the x-axis.
2. A uniform electric field \(\vec{E} = E_0 \hat{j}\) is present along the y-axis. An electric field exerts a force on a charged particle. For an electron with charge \(-e\), the force \(\vec{F}\) it experiences is given by: \(\vec{F} = -e \vec{E} = -e E_0 \hat{j}\).
3. This force causes an acceleration \(\vec{a}\) in the y-direction (since force and acceleration are directly proportional), given by: \(\vec{a} = \frac{\vec{F}}{m} = -\frac{e E_0}{m} \hat{j}\), where \(m\) is the mass of the electron.
4. In the x-direction, there is no force acting on the electron (assuming no magnetic field or other forces), so it continues to move with constant velocity \(v_0 \hat{i}\).
5. The motion in the y-direction is influenced by the constant acceleration \(\vec{a}\). According to the equations of motion, the displacement in the y-direction \(y\) is given by: \(y = \frac{1}{2} a t^2 = -\frac{1}{2} \frac{e E_0}{m} t^2\)
6. Combining the motion in both x and y directions, the trajectory followed by the electron will be described by: \(y = -\frac{e E_0}{2m} t^2\), which is a parabolic path along the plane described by \(x\) and \(y\) coordinates.

Therefore, under the influence of the given uniform electric field, the electron describes a parabolic path.