Problem 22 - A Complete Resource Book in Physics for JEE Main

Question

The magnitude of electric field intensity \(E\) is such that, an electron of mass \(m\) and charge \(e\) placed in it would experience an electrical force equal to its weight is given by (A) \(m g e\) (B) \(\frac{m g}{e}\) (C) \(\frac{e}{m g}\) (D) \(\frac{e^{2}}{m^{2}} g\)

Step-by-Step Solution

Answer

The short answer is: The correct expression for the magnitude of the electric field intensity E is (B) \(\frac{m g}{e}\).

Step 1: Identify the forces on the electron

The electron experiences two forces: the gravitational force (its weight) and the electrical force due to the electric field E.

Step 2: Write expressions for the forces

Gravitational Force (Weight) is given by: Weight = m * g where m is the mass of the electron, and g is the acceleration due to gravity. The electrical force on the electron is given by: Electrical Force = e * E where e is the charge of the electron and E is the magnitude of the electric field intensity.

Step 3: Set up an equation relating the forces

Since it is given that the electrical force on the electron is equal to its weight, we can write: Electrical Force = Weight e * E = m * g

Step 4: Solve for the electric field intensity E

Solving the equation for E, we get: E = \(\frac{m*g}{e}\)

Step 5: Choose the correct expression from the given options

Comparing our calculated expression for E to the choices given in the problem, we find that the correct answer is: (B) \(\frac{m g}{e}\)