Let `E_1(r)`, `E_2(r)` and `E_3(r)` be the respectively electric field at a distance r from a point charge `Q`, an infinitely long wire with constant linear charge density `lambda`, and an infinite plane with uniform surface charge density `sigma`. If `E_1(r_0)=E_2(r_0)=E_3(r_0)` at a given distance `r_0`, then

Options

A. `Q = 4 sigma r_(0)^(2)`
B. `r_(0) = (lambda)/(2 pi r)`
C. `E_(1) (r_(0)//2) = 2E_(2) (r_(0)//2)`
D. `E_(2) (r_(0) //2) = 4E_(a) (r_(0)//2)`

Explanation

To solve the problem, we need to find the relationship between the electric fields produced by a point charge, an infinitely long wire, and an infinite plane at a given distance \( r_0 \).

### Step 1: Write the expressions for the electric fields

1. **Electric field due to a point charge \( Q \)** at a distance \( r_0 \):
\[ E_1(r_0) = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r_0^2} \]

2. **Electric field due to an infinitely long wire with linear charge density \( \lambda \)** at a distance \( r_0 \):
\[ E_2(r_0) = \frac{1}{4\pi \epsilon_0} \cdot \frac{2\lambda}{r_0} \]

3. **Electric field due to an infinite plane with surface charge density \( \sigma \)** (which is constant and does not depend on distance):
\[ E_3 = \frac{\sigma}{2\epsilon_0} \]

### Step 2: Set the electric fields equal to each other

Given that \( E_1(r_0) = E_2(r_0) = E_3 \), we can equate these expressions.

1. **Equate \( E_1 \) and \( E_2 \)**:
\[ \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r_0^2} = \frac{1}{4\pi \epsilon_0} \cdot \frac{2\lambda}{r_0} \]
Simplifying gives:
\[ \frac{Q}{r_0^2} = \frac{2\lambda}{r_0} \]
Multiplying both sides by \( r_0^2 \):
\[ Q = 2\lambda r_0 \]

2. **Equate \( E_2 \) and \( E_3 \)**:
\[ \frac{1}{4\pi \epsilon_0} \cdot \frac{2\lambda}{r_0} = \frac{\sigma}{2\epsilon_0} \]
Simplifying gives:
\[ \frac{2\lambda}{r_0} = \frac{2\sigma}{4\pi} \]
Multiplying both sides by \( r_0 \):
\[ 2\lambda = \frac{2\sigma r_0}{4\pi} \]
Simplifying gives:
\[ \lambda = \frac{\sigma r_0}{4\pi} \]

### Step 3: Substitute \( \lambda \) into the expression for \( Q \)

Now substitute the expression for \( \lambda \) back into the equation for \( Q \):
\[ Q = 2\left(\frac{\sigma r_0}{4\pi}\right)r_0 = \frac{\sigma r_0^2}{2\pi} \]

### Summary of Results

We have derived the following relationships:
1. \( Q = 2\lambda r_0 \)
2. \( \lambda = \frac{\sigma r_0}{4\pi} \)
3. \( Q = \frac{\sigma r_0^2}{2\pi} \)

### Conclusion

The relationships we derived show how the charge \( Q \), the linear charge density \( \lambda \), and the surface charge density \( \sigma \) are interconnected when the electric fields are equal at a distance \( r_0 \).