Two points A and B located in diametrically opposite directions of a point charge of +2`muC` at distances 2.0 m and 1.0 m respectively from it. Determine the potential difference `V_(A)-V_(B)`

Options

A. `3xx10^(3)V`
B. `6xx10^(4)V`
C. `-9xx10^(3)V`
D. `-3xx10^(3)V`

Explanation

To determine the potential difference \( V_A - V_B \) between the points A and B located at distances of 2.0 m and 1.0 m respectively from a point charge of \( +2 \, \mu C \), we can follow these steps:

### Step 1: Understand the formula for electric potential
The electric potential \( V \) due to a point charge \( Q \) at a distance \( r \) is given by the formula:
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

### Step 2: Calculate the potential at point A
For point A, which is at a distance \( R_A = 2.0 \, m \):
\[ V_A = \frac{kQ}{R_A} = \frac{(8.99 \times 10^9) \times (2 \times 10^{-6})}{2} \]
Calculating this:
\[ V_A = \frac{(8.99 \times 10^9) \times (2 \times 10^{-6})}{2} = \frac{(8.99 \times 2)}{2} \times 10^3 = 8.99 \times 10^3 \, V \]

### Step 3: Calculate the potential at point B
For point B, which is at a distance \( R_B = 1.0 \, m \):
\[ V_B = \frac{kQ}{R_B} = \frac{(8.99 \times 10^9) \times (2 \times 10^{-6})}{1} \]
Calculating this:
\[ V_B = (8.99 \times 10^9) \times (2 \times 10^{-6}) = 17.98 \times 10^3 \, V \]

### Step 4: Calculate the potential difference \( V_A - V_B \)
Now, we can find the potential difference:
\[ V_A - V_B = 8.99 \times 10^3 - 17.98 \times 10^3 \]
Calculating this:
\[ V_A - V_B = 8.99 \times 10^3 - 17.98 \times 10^3 = -8.99 \times 10^3 \, V \]

### Final Result
Thus, the potential difference \( V_A - V_B \) is:
\[ V_A - V_B = -8.99 \times 10^3 \, V \]