Question

Electric field at a distance ‘r’ from an infinitely long uniformly charged straight conductor, having linear charge density λ is E₁. Another uniformly charged conductor having same linear charge density λ is bent into a semicircle of radius ‘r’. The electric field at its centre is E₂ . Then

• A. \(E_2=\frac{E_1}{r}\)
• B. E₁ = E₂
• C. E₁ = πrE₂
• D. E₂ = πrE₁

Answer 1

The Correct Option is B

Given problem: 

• Electric field at a distance \( r \) from an infinitely long uniformly charged straight conductor with linear charge density \( \lambda \) is \( E_1 \).
• Another uniformly charged conductor with the same linear charge density \( \lambda \) is bent into a semicircle of radius \( r \). The electric field at its center is \( E_2 \).

We need to find the relationship between \( E_1 \) and \( E_2 \).

Step 1: Electric Field due to an Infinitely Long Uniformly Charged Straight Conductor

The electric field \( E_1 \) at a distance \( r \) from an infinitely long uniformly charged straight conductor with linear charge density \( \lambda \) is given by:

\( E_1 = \frac{\lambda}{2\pi \epsilon_0 r} \)

Step 2: Electric Field due to a Semicircular Conductor

Consider a semicircular conductor of radius \( r \) with linear charge density \( \lambda \). We need to find the electric field \( E_2 \) at the center of this semicircle.

To do this, we can use the principle of symmetry and integration. The semicircle can be divided into small segments, each contributing to the electric field at the center. Due to symmetry, the horizontal components of the electric field from each segment will cancel out, leaving only the vertical components.

The electric field \( dE \) due to a small segment \( dq \) of the semicircle at an angle \( \theta \) from the vertical is given by:

\( dE = \frac{dq}{4\pi \epsilon_0 r^2} \)

where \( dq = \lambda \, ds \) and \( ds = r \, d\theta \).

Thus,

\( dE = \frac{\lambda r \, d\theta}{4\pi \epsilon_0 r^2} = \frac{\lambda \, d\theta}{4\pi \epsilon_0 r} \)

The vertical component of \( dE \) is:

\( dE_y = dE \sin(\theta) = \frac{\lambda \sin(\theta) \, d\theta}{4\pi \epsilon_0 r} \)

Integrating \( dE_y \) from \( \theta = 0 \) to \( \theta = \pi \):

\( E_2 = \int_0^\pi \frac{\lambda \sin(\theta) \, d\theta}{4\pi \epsilon_0 r} \)

\( E_2 = \frac{\lambda}{4\pi \epsilon_0 r} \int_0^\pi \sin(\theta) \, d\theta \)

\( E_2 = \frac{\lambda}{4\pi \epsilon_0 r} \left[ -\cos(\theta) \right]_0^\pi \)

\( E_2 = \frac{\lambda}{4\pi \epsilon_0 r} \left( -\cos(\pi) + \cos(0) \right) \)

\( E_2 = \frac{\lambda}{4\pi \epsilon_0 r} \left( 1 + 1 \right) \)

\( E_2 = \frac{\lambda}{4\pi \epsilon_0 r} \cdot 2 \)

\( E_2 = \frac{\lambda}{2\pi \epsilon_0 r} \)

Step 3: Comparing \( E_1 \) and \( E_2 \)

From the calculations, we see that:

\( E_1 = \frac{\lambda}{2\pi \epsilon_0 r} \)

\( E_2 = \frac{\lambda}{2\pi \epsilon_0 r} \)

Thus, \( E_1 = E_2 \).

Therefore, the correct answer is:

\( \boxed{E_1 = E_2} \)

Answer 2

1. Electric Field due to an Infinitely Long Straight Conductor (\(E_1\)):
The electric field \(E\) at a perpendicular distance \(r\) from an infinitely long straight conductor with uniform linear charge density \(\lambda\) is given by the formula (derived using Gauss's Law): \[ E_1 = \frac{\lambda}{2\pi\varepsilon_0 r} \] Where \(\varepsilon_0\) is the permittivity of free space.

2. Electric Field at the Center of a Uniformly Charged Semicircle (\(E_2\)):
Consider a uniformly charged semicircle of radius \(r\) and linear charge density \(\lambda\). We need to find the electric field at its center (the center of curvature).
Let's consider a small element of the semicircle of length \(dl\). This element carries a charge \(dq = \lambda dl\). If we define the angle \(\theta\) from one end of the semicircle, then \(dl = r d\theta\).
The electric field \(dE\) produced by this element \(dq\) at the center has magnitude: \[ dE = \frac{dq}{4\pi\varepsilon_0 r^2} = \frac{\lambda dl}{4\pi\varepsilon_0 r^2} = \frac{\lambda (r d\theta)}{4\pi\varepsilon_0 r^2} = \frac{\lambda}{4\pi\varepsilon_0 r} d\theta \] This field points radially away from the element \(dq\). Due to symmetry, the components of the electric field parallel to the diameter of the semicircle cancel out. Only the components perpendicular to the diameter (along the axis of symmetry) add up.
Let the axis of symmetry be the y-axis, passing through the center and perpendicular to the diameter. The component of \(dE\) along the y-axis is \(dE_y = dE \sin\theta\), where \(\theta\) is measured from the end of the diameter.
We integrate from \(\theta = 0\) to \(\theta = \pi\): \[ E_2 = \int dE_y = \int_{0}^{\pi} \left( \frac{\lambda}{4\pi\varepsilon_0 r} \right) \sin\theta \, d\theta \] \[ E_2 = \frac{\lambda}{4\pi\varepsilon_0 r} \int_{0}^{\pi} \sin\theta \, d\theta \] \[ E_2 = \frac{\lambda}{4\pi\varepsilon_0 r} [-\cos\theta]_{0}^{\pi} \] \[ E_2 = \frac{\lambda}{4\pi\varepsilon_0 r} [-\cos(\pi) - (-\cos(0))] \] \[ E_2 = \frac{\lambda}{4\pi\varepsilon_0 r} [-(-1) - (-1)] \] \[ E_2 = \frac{\lambda}{4\pi\varepsilon_0 r} [1 + 1] \] \[ E_2 = \frac{\lambda}{4\pi\varepsilon_0 r} \times 2 \] \[ E_2 = \frac{2\lambda}{4\pi\varepsilon_0 r} = \frac{\lambda}{2\pi\varepsilon_0 r} \]

3. Comparison:
We found: \[ E_1 = \frac{\lambda}{2\pi\varepsilon_0 r} \] \[ E_2 = \frac{\lambda}{2\pi\varepsilon_0 r} \] Therefore, \(E_1 = E_2\).

Answer: The correct option is (B) \(E_1 = E_2\).