Along the x-axis, three charges `q/2,-q` and `q/2` are placed at `x=0,x=a` and `x=2a` respectively. The resultant electric potential at a point P located at a distance r from the charge `-q`, `(rgtgta)` is : ( `epsilon_(0)` is the permittivity of free space)

Options

A. `(aq)/(4piepsilon_(0)r^(2))`
B. `(qa^(2))/(4piepsilon_(0)r^(3))`
C. `(q(a^(2)//4))/(4piepsilon_(0)r^(2))`
D. `(q)/(4piepsilon_(0)r^(2))`

Explanation

To find the resultant electric potential at point P located at a distance \( r \) from the charge \( -q \), we will follow these steps:

### Step 1: Identify the positions of the charges
- The charges are placed as follows:
- Charge \( \frac{q}{2} \) at \( x = 0 \)
- Charge \( -q \) at \( x = a \)
- Charge \( \frac{q}{2} \) at \( x = 2a \)

### Step 2: Determine the distance from point P to each charge
- Point P is located at a distance \( r \) from the charge \( -q \) at \( x = a \).
- Therefore, the distances from point P to each charge are:
- Distance to charge \( \frac{q}{2} \) at \( x = 0 \): \( d_1 = r + a \)
- Distance to charge \( -q \) at \( x = a \): \( d_2 = r \)
- Distance to charge \( \frac{q}{2} \) at \( x = 2a \): \( d_3 = r - a \)

### Step 3: Write the expression for electric potential
- The electric potential \( V \) due to a point charge is given by:
\[ V = k \frac{q}{r} \]
where \( k = \frac{1}{4\pi \epsilon_0} \).

- The total electric potential \( V_P \) at point P due to the three charges is the sum of the potentials due to each charge:
\[ V_P = V_1 + V_2 + V_3 \]
where:
- \( V_1 = k \frac{\frac{q}{2}}{d_1} = k \frac{\frac{q}{2}}{r + a} \)
- \( V_2 = k \frac{-q}{d_2} = k \frac{-q}{r} \)
- \( V_3 = k \frac{\frac{q}{2}}{d_3} = k \frac{\frac{q}{2}}{r - a} \)

### Step 4: Substitute the distances into the potential expression
- Substitute the distances into the expression for \( V_P \):
\[ V_P = k \left( \frac{\frac{q}{2}}{r + a} - \frac{q}{r} + \frac{\frac{q}{2}}{r - a} \right) \]

### Step 5: Combine the terms
- To combine the terms, we need a common denominator, which is \( (r + a)(r)(r - a) \):
\[ V_P = k \left( \frac{\frac{q}{2} r (r - a) - q (r + a)(r - a) + \frac{q}{2} r (r + a)}{(r + a)(r)(r - a)} \right) \]

### Step 6: Simplify the expression
- After simplifying the numerator, we can factor out terms and reduce the expression. Given that \( r \) is much greater than \( a \), we can neglect terms involving \( a^2 \) compared to \( r^2 \):
\[ V_P \approx \frac{q a^2}{4 \pi \epsilon_0 r^3} \]

### Final Result
Thus, the resultant electric potential at point P is:
\[ V_P = \frac{q a^2}{4 \pi \epsilon_0 r^3} \]