Question

The electric field lines on the left have twice the separation on those on the right as shown in figure. If the magnitude of the field at $A$ is $40\,Vm^{-1}$, what is the force on $20\,\mu \,C$ charge kept at $B$?

• A. $4 \times 10^{-4} Vm^{-1}$
• B. $8 \times 10^{-4} Vm^{-1}$
• C. $16 \times 10^{-4} Vm^{-1}$
• D. $1 \times 10^{-4} Vm^{-1}$

Answer 1

The Correct Option is A

Problem Analysis:

1. The problem provides a diagram of electric field lines.
2. It states that the separation between the field lines on the left (region B) is twice the separation between the field lines on the right (region A). Let the separation at A be $d_A$ and at B be $d_B$. So, $d_B = 2 d_A$.
3. The magnitude of the electric field at point A is given as $E_A = 40 \, \text{Vm}^{-1}$.
4. We need to find the force $F_B$ on a charge $q = 20 \, \mu\text{C} = 20 \times 10^{-6} \, \text{C}$ placed at point B.

Key Concept:

The density of electric field lines (number of lines per unit area perpendicular to the lines) is proportional to the magnitude of the electric field strength. Therefore, where the lines are closer together, the field is stronger, and where they are farther apart, the field is weaker. Specifically, the electric field strength $E$ is inversely proportional to the separation $d$ between the lines (assuming a uniform distribution in the perpendicular dimension shown):

$E \propto \frac{1}{d}$

This implies that the ratio of electric fields at two points is inversely proportional to the ratio of the separations at those points:

$\frac{E_B}{E_A} = \frac{d_A}{d_B}$

Calculations:

1. Use the relationship between field strength and separation:

$\frac{E_B}{E_A} = \frac{d_A}{d_B}$

Substitute $d_B = 2 d_A$:

$\frac{E_B}{E_A} = \frac{d_A}{2 d_A} = \frac{1}{2}$

1. Calculate the electric field strength at B ($E_B$):

$E_B = \frac{1}{2} E_A$

Given $E_A = 40 \, \text{Vm}^{-1}$:

$E_B = \frac{1}{2} \times 40 \, \text{Vm}^{-1} = 20 \, \text{Vm}^{-1}$

1. Calculate the force on the charge $q$ at point B ($F_B$) using the formula $F = qE$:

$F_B = q \times E_B$

Substitute $q = 20 \, \mu\text{C} = 20 \times 10^{-6} \, \text{C}$ and $E_B = 20 \, \text{Vm}^{-1}$ (Note: $1 \, \text{Vm}^{-1} = 1 \, \text{N/C}$):

$F_B = (20 \times 10^{-6} \, \text{C}) \times (20 \, \text{N/C})$

$F_B = 400 \times 10^{-6} \, \text{N}$

$F_B = 4 \times 10^2 \times 10^{-6} \, \text{N}$

$F_B = 4 \times 10^{-4} \, \text{N}$

The final answer is (A): \(4 \times 10^{-4} N\).

Answer 2

We are given that the electric field at point A is 40 Vm⁻¹, and the electric field lines on the left are twice as far apart as those on the right. This means that the electric field at point B will be half of the electric field at point A. The force on a charge \( q \) in an electric field \( E \) is given by: \[ F = qE \] Given that the charge \( q = 20 \, \mu C = 20 \times 10^{-6} \, C \) and the electric field at point B is half of that at A, we can calculate the electric field at B as: \[ E_B = \frac{40}{2} = 20 \, \text{Vm}^{-1} \] Now, the force at point B is: \[ F = 20 \times 10^{-6} \times 20 = 4 \times 10^{-4} \, \text{N} \] Thus, the force on the 20 μC charge at point B is \( 4 \times 10^{-4} \, \text{N} \).