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

Explanation

To find the resultant electric potential at point P due to the three charges placed along the x-axis, we will follow these steps:

### Step 1: Identify the positions of the charges
The charges are placed at the following positions:
- 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
Let the point P be located at a distance \( r \) from the charge \( -q \) (which is at \( x = a \)). Therefore, the distances from point P to the other charges 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 the electric potential due to each charge
The electric potential \( V \) due to a point charge is given by the formula:
\[ V = \frac{kQ}{r} \]
where \( k = \frac{1}{4\pi\epsilon_0} \) is Coulomb's constant.

Thus, the potentials due to each charge at point P are:
- Due to charge \( \frac{q}{2} \) at \( x = 0 \):
\[ V_1 = \frac{k \cdot \frac{q}{2}}{r + a} \]
- Due to charge \( -q \) at \( x = a \):
\[ V_2 = \frac{-kq}{r} \]
- Due to charge \( \frac{q}{2} \) at \( x = 2a \):
\[ V_3 = \frac{k \cdot \frac{q}{2}}{r - a} \]

### Step 4: Calculate the total electric potential at point P
The total electric potential \( V \) at point P is the sum of the potentials due to all three charges:
\[ V = V_1 + V_2 + V_3 \]
Substituting the expressions from Step 3:
\[ V = \frac{k \cdot \frac{q}{2}}{r + a} - \frac{kq}{r} + \frac{k \cdot \frac{q}{2}}{r - a} \]

### Step 5: Simplify the expression
To simplify, we can factor out \( kq \):
\[ V = kq \left( \frac{1/2}{r + a} - \frac{1}{r} + \frac{1/2}{r - a} \right) \]
Now, we need to find a common denominator for the terms inside the parentheses:
The common denominator is \( (r + a)(r)(r - a) \).

### Step 6: Combine the fractions
Rewriting each term with the common denominator:
\[ \frac{1/2 \cdot r(r - a) - r(r + a) + 1/2 \cdot r(r + a)}{(r + a)(r)(r - a)} \]
Now, simplifying the numerator:
1. The first term: \( \frac{1}{2} r^2 - \frac{1}{2} ra \)
2. The second term: \( -r^2 - ra \)
3. The third term: \( \frac{1}{2} r^2 + \frac{1}{2} ra \)

Combining these gives:
\[ \frac{(1/2 + 1/2 - 1) r^2 + (-1/2) ra}{(r + a)(r)(r - a)} = \frac{-\frac{1}{2} r^2 - \frac{1}{2} ra}{(r + a)(r)(r - a)} \]

### Step 7: Final expression for potential
Thus, the total potential \( V \) becomes:
\[ V = kq \cdot \frac{-\frac{1}{2} (r^2 + ra)}{(r + a)(r)(r - a)} \]

### Step 8: Simplifying further
As \( r \) becomes very large compared to \( a \), we can approximate the potential:
\[ V \approx \frac{kq a^2}{r^3} \]
Thus, the final expression for the electric potential at point P is:
\[ V = \frac{1}{4\pi\epsilon_0} \cdot \frac{q a^2}{r^3} \]