Three charged particles having charges q, -2q & q are placed in a line at points (-a, 0), (0,0) & (a , 0) respectively. The expression for electric potential at P(r, 0 ) for r `gt gt ` a is

Options

A. `(1)/(4 pi epsilon_(0))(qa^(2))/(r^(4))`
B. `(1)/(4 pi epsilon_(0))(2qa^(2))/(r^(3))`
C. `(1)/(4 pi epsilon_(0))(4qa^(2))/(r^(2))`
D. `(1)/(4 pi epsilon_(0))(8qa^(2))/(r )`

Answer

To find the electric potential at point P(r, 0) due to three charged particles with charges q, -2q, and q placed at (-a, 0), (0, 0), and (a, 0) respectively, we can follow these steps:

### Step 1: Identify the positions of the charges and the point P
The charges are located at:
- Charge q at (-a, 0)
- Charge -2q at (0, 0)
- Charge q at (a, 0)

The point P where we want to find the electric potential is at (r, 0).

### Step 2: Write the expression for electric potential due to each charge
The electric potential V due to a point charge Q at a distance d is given by the formula:
\[ V = k \frac{Q}{d} \]
where \( k = \frac{1}{4\pi \epsilon_0} \).

#### For charge q at (-a, 0):
The distance from P to this charge is:
\[ d_1 = r + a \]
Thus, the potential due to this charge is:
\[ V_1 = k \frac{q}{r + a} \]

#### For charge -2q at (0, 0):
The distance from P to this charge is:
\[ d_2 = r \]
Thus, the potential due to this charge is:
\[ V_2 = k \frac{-2q}{r} \]

#### For charge q at (a, 0):
The distance from P to this charge is:
\[ d_3 = r - a \]
Thus, the potential due to this charge is:
\[ V_3 = k \frac{q}{r - a} \]

### Step 3: Write the total electric potential at point P
The total electric potential V at point P is the sum of the potentials due to each charge:
\[ V = V_1 + V_2 + V_3 \]
Substituting the expressions we derived:
\[ V = k \frac{q}{r + a} - k \frac{2q}{r} + k \frac{q}{r - a} \]

### Step 4: Factor out k and simplify the expression
Factoring out \( k \):
\[ V = k \left( \frac{q}{r + a} - \frac{2q}{r} + \frac{q}{r - a} \right) \]

### Step 5: Find a common denominator and simplify further
The common denominator for the fractions is \( (r + a)(r)(r - a) \). Rewriting the expression:
\[ V = kq \left( \frac{r(r - a) + r(r + a) - 2(r + a)(r - a)}{(r + a)(r)(r - a)} \right) \]

### Step 6: Simplify the numerator
Expanding the terms in the numerator:
1. \( r(r - a) = r^2 - ar \)
2. \( r(r + a) = r^2 + ar \)
3. \( -2(r + a)(r - a) = -2(r^2 - a^2) = -2r^2 + 2a^2 \)

Combining these:
\[ r^2 - ar + r^2 + ar - 2r^2 + 2a^2 = 2a^2 \]

Thus, the numerator simplifies to \( 2a^2 \).

### Step 7: Write the final expression for electric potential
Now substituting back into the potential expression:
\[ V = kq \frac{2a^2}{(r + a)(r)(r - a)} \]

### Step 8: Consider the limit as \( r \gg a \)
In the limit where \( r \) is much greater than \( a \), we can approximate:
- \( r + a \approx r \)
- \( r - a \approx r \)

Thus, the expression simplifies to:
\[ V \approx kq \frac{2a^2}{r^3} \]

### Final Expression
The final expression for the electric potential at point P is:
\[ V = \frac{2kqa^2}{r^3} \]