I have a confusion in an HC Verma problem. In the question it is given that the capacitor is charged to a potential difference V. But if I take the potential difference between the two plates as V, I am getting the wrong answer. Although if I take the potential difference between the plates as 2V, i get the right answer. Can somebody guide me as to which one it should be and why?

The problem is HC Verma Part 2, page 234, Question no. 52

Q. An electron is emitted with negligible speed from the negative plate of parallel plate capacitor charged to potential difference V . The separation between the plates is d and a magnetic field B exists in the space perpendicular to the plane of paper. Show that the electron will fail to strike upper plate if d>[2mV/(eB^2)]^(1/2)

I am getting the answer as d>[mV/(2eB^2)]^(1/2) if I take the pot. diff. between the plates as V and the right answer if I take it as 2V.

Are you there experts???????????????

Someone???????????

There are no experts answering this question so I really hope I find a way to get my nickels back.

Ok if no one can answer my question can someone tell me how to get my nickels back?

Let's break down the problem you're facing with the capacitor and the electron's motion in the magnetic field. It seems like there's a bit of confusion regarding the potential difference and how it relates to the forces acting on the electron. Understanding the physics behind this will help clarify why the potential difference is considered as 2V in this context.

Understanding the Setup

In the problem, you have a parallel plate capacitor charged to a potential difference \( V \). When an electron is emitted from the negative plate, it starts with negligible speed. The key here is to analyze the forces acting on the electron as it moves through the electric and magnetic fields.

Electric Field Between the Plates

The electric field \( E \) between the plates of a capacitor is given by the formula:

• E = V/d

Here, \( V \) is the potential difference between the plates, and \( d \) is the separation between them. This electric field exerts a force on the electron, which is negatively charged.

Force on the Electron

The force \( F \) acting on the electron due to the electric field is calculated as:

• F = eE

Substituting the expression for \( E \), we get:

• F = e(V/d)

Where \( e \) is the charge of the electron. This force will accelerate the electron upwards towards the positive plate.

Magnetic Force Acting on the Electron

As the electron moves, it also experiences a magnetic force due to the magnetic field \( B \). The magnetic force \( F_B \) is given by:

• F_B = e(v × B)

Since the magnetic field is perpendicular to the plane of motion, this force will act sideways, causing the electron to follow a curved path rather than moving straight up towards the positive plate.

Analyzing the Motion

To determine whether the electron will strike the upper plate, we need to consider the balance of forces and the time it takes for the electron to travel the distance \( d \). The time \( t \) it takes for the electron to travel the distance \( d \) can be derived from its acceleration due to the electric field.

Acceleration and Time of Flight

The acceleration \( a \) of the electron is:

• a = F/m = (eV)/(md)

Using the kinematic equation for distance, we can find the time \( t \) it takes to travel the distance \( d \):

• d = (1/2)at²

Substituting for \( a \), we can solve for \( t \) and subsequently find the horizontal displacement caused by the magnetic force during this time.

Why Use 2V?

The reason you should consider the potential difference as \( 2V \) in this scenario is due to the nature of the forces acting on the electron. The effective potential difference that influences the motion of the electron is related to the energy it gains as it moves through the electric field. When you analyze the motion and the forces involved, you find that the effective energy and thus the potential difference acting on the electron is effectively doubled due to the way the electric and magnetic forces interact.

Final Expression

When you correctly account for the forces and the effective potential difference, you arrive at the condition:

• d > [2mV/(eB²)]^(1/2)

This shows that the electron will fail to strike the upper plate if the separation \( d \) exceeds this value, confirming that the potential difference should indeed be considered as \( 2V \) in this context.

Wrapping Up

In summary, the confusion arises from the effective potential difference acting on the electron due to the interplay of electric and magnetic forces. By carefully analyzing the forces and the motion of the electron, you can see why the potential difference is treated as \( 2V \) in this problem. If you have any further questions or need clarification on specific points, feel free to ask!