Two point charges `-q` and `+q` are located at points `(0,0-a)` and `(0,0,a)` respectively. The electric potential at point `(0,0,z)`is `(Z gt a)`

Options

A. `(qa)/(4pi_(0)z^(2))`
B. `(q)/(4pi epsilon_(0)a)`
C. `(2qa)/(4pi epsilon_(0)(z^(2)-a^(2)))`
D. `(2qa)/(4piepsilon_(0)(z^(2)+a^(2)))`

Answer

To find the electric potential at the point (0, 0, z) due to the two point charges -q and +q located at (0, 0, -a) and (0, 0, a) respectively, we can follow these steps:

### Step 1: Identify the distances from the point of interest to each charge
The distance from the point (0, 0, z) to the charge -q at (0, 0, -a) is:
\[ r_1 = z + a \]
The distance from the point (0, 0, z) to the charge +q at (0, 0, a) is:
\[ r_2 = z - a \]

### Step 2: Write the expression for electric potential due to each charge
The electric potential \( V \) at a point due to a point charge \( Q \) is given by:
\[ V = k \frac{Q}{r} \]
where \( k \) is Coulomb's constant.

Thus, the potential at (0, 0, z) due to charge -q is:
\[ V_{-q} = k \frac{-q}{z + a} \]
And the potential at (0, 0, z) due to charge +q is:
\[ V_{+q} = k \frac{q}{z - a} \]

### Step 3: Calculate the total electric potential at (0, 0, z)
The total electric potential \( V \) at the point (0, 0, z) is the sum of the potentials due to both charges:
\[ V = V_{-q} + V_{+q} \]
Substituting the expressions we found:
\[ V = k \frac{-q}{z + a} + k \frac{q}{z - a} \]

### Step 4: Combine the fractions
To combine the fractions, we need a common denominator:
\[ V = kq \left( \frac{-1}{z + a} + \frac{1}{z - a} \right) \]
The common denominator is \( (z + a)(z - a) \):
\[ V = kq \left( \frac{-(z - a) + (z + a)}{(z + a)(z - a)} \right) \]
Simplifying the numerator:
\[ V = kq \left( \frac{-z + a + z + a}{(z + a)(z - a)} \right) \]
\[ V = kq \left( \frac{2a}{(z + a)(z - a)} \right) \]

### Step 5: Simplify the expression
The denominator can be simplified using the difference of squares:
\[ (z + a)(z - a) = z^2 - a^2 \]
Thus, the potential becomes:
\[ V = \frac{2kaq}{z^2 - a^2} \]

### Step 6: Substitute the value of k
Since \( k = \frac{1}{4 \pi \epsilon_0} \), we can substitute this into the equation:
\[ V = \frac{2q}{4 \pi \epsilon_0} \cdot \frac{a}{z^2 - a^2} \]
This gives us the final expression for the electric potential at the point (0, 0, z).

### Final Answer
The electric potential at the point (0, 0, z) is:
\[ V = \frac{2qa}{4 \pi \epsilon_0 (z^2 - a^2)} \]