Ratio of electric and magnetic field due to moving point charge if its speed is `4.5xx10^(5)m//s`

Options

A. `2xx10^(11)`
B. `3xx10^(11)`
C. `2xx10^(8)`
D. `3xx10^(12)`

Explanation

To solve the problem of finding the ratio of the electric field (E) to the magnetic field (B) due to a moving point charge, we can follow these steps:

### Step 1: Write the formulas for electric and magnetic fields
The electric field (E) due to a point charge \( q \) at a distance \( r \) is given by:
\[ E = \frac{kq}{r^2} \]
where \( k \) is Coulomb's constant, \( k = \frac{1}{4\pi \epsilon_0} \).

The magnetic field (B) due to a moving charge is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{qv}{r^2} \]
where \( v \) is the velocity of the charge and \( \mu_0 \) is the permeability of free space.

### Step 2: Set up the ratio \( \frac{E}{B} \)
Now, we need to find the ratio \( \frac{E}{B} \):
\[ \frac{E}{B} = \frac{\frac{kq}{r^2}}{\frac{\mu_0}{4\pi} \cdot \frac{qv}{r^2}} \]

### Step 3: Simplify the ratio
In the ratio, \( r^2 \) and \( q \) will cancel out:
\[ \frac{E}{B} = \frac{k}{\frac{\mu_0}{4\pi} \cdot v} \]
This can be rewritten as:
\[ \frac{E}{B} = \frac{4\pi k}{\mu_0 v} \]

### Step 4: Substitute \( k \) and simplify further
Substituting \( k = \frac{1}{4\pi \epsilon_0} \):
\[ \frac{E}{B} = \frac{4\pi \cdot \frac{1}{4\pi \epsilon_0}}{\mu_0 v} = \frac{1}{\epsilon_0 \mu_0 v} \]

### Step 5: Use the relationship between \( \epsilon_0 \), \( \mu_0 \), and the speed of light
We know that:
\[ c^2 = \frac{1}{\epsilon_0 \mu_0} \]
Thus, we can substitute:
\[ \frac{E}{B} = \frac{c^2}{v} \]

### Step 6: Substitute the values
Given \( v = 4.5 \times 10^5 \, \text{m/s} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \]
Now substituting into the ratio:
\[ \frac{E}{B} = \frac{9 \times 10^{16}}{4.5 \times 10^5} \]

### Step 7: Calculate the final value
Now, we can perform the division:
\[ \frac{E}{B} = \frac{9}{4.5} \times 10^{16 - 5} = 2 \times 10^{11} \]

### Final Answer
Thus, the ratio of the electric field to the magnetic field due to the moving point charge is:
\[ \frac{E}{B} = 2 \times 10^{11} \]