To find the electric potential at a point in the first quadrant due to the three charges located at (0, a), (0, 0), and (0, -a), we will follow these steps:
### Step 1: Identify the Charges and Their Positions
We have three charges:
- Charge +q at (0, a)
- Charge +q at (0, 0)
- Charge -q at (0, -a)
### Step 2: Determine the Position of the Point of Interest
Let the point of interest in the first quadrant be denoted as P. The position vector of point P makes an angle θ with the y-axis and is at a distance r from the origin.
### Step 3: Calculate the Distance from Each Charge to Point P
Using trigonometry, we can express the coordinates of point P:
- x-coordinate: \( r \sin(\theta) \)
- y-coordinate: \( r \cos(\theta) \)
Now, we can calculate the distances from each charge to point P:
1. Distance from charge +q at (0, a):
\[
d_1 = \sqrt{(r \sin(\theta) - 0)^2 + (r \cos(\theta) - a)^2} = \sqrt{r^2 \sin^2(\theta) + (r \cos(\theta) - a)^2}
\]
2. Distance from charge +q at (0, 0):
\[
d_2 = \sqrt{(r \sin(\theta) - 0)^2 + (r \cos(\theta) - 0)^2} = \sqrt{r^2 \sin^2(\theta) + r^2 \cos^2(\theta)} = r
\]
3. Distance from charge -q at (0, -a):
\[
d_3 = \sqrt{(r \sin(\theta) - 0)^2 + (r \cos(\theta) + a)^2} = \sqrt{r^2 \sin^2(\theta) + (r \cos(\theta) + a)^2}
\]
### Step 4: Calculate the Electric Potential Due to Each Charge
The electric potential \( V \) at point P due to a point charge \( Q \) is given by:
\[
V = \frac{kQ}{d}
\]
where \( k \) is Coulomb's constant.
1. Potential due to charge +q at (0, a):
\[
V_1 = \frac{kq}{d_1}
\]
2. Potential due to charge +q at (0, 0):
\[
V_2 = \frac{kq}{d_2} = \frac{kq}{r}
\]
3. Potential due to charge -q at (0, -a):
\[
V_3 = \frac{-kq}{d_3}
\]
### Step 5: Total Electric Potential at Point P
The total electric potential \( V_P \) at point P is the sum of the potentials due to each charge:
\[
V_P = V_1 + V_2 + V_3 = \frac{kq}{d_1} + \frac{kq}{r} - \frac{kq}{d_3}
\]
### Step 6: Simplify the Expression
Since \( r \gg a \), we can approximate \( d_1 \) and \( d_3 \) using binomial expansion or Taylor series:
- For \( d_1 \) and \( d_3 \), we can expand and simplify, leading to:
\[
d_1 \approx r - a \cos(\theta) \quad \text{and} \quad d_3 \approx r + a \cos(\theta)
\]
### Step 7: Substitute and Finalize the Expression
Substituting these approximations back into the expression for \( V_P \):
\[
V_P \approx \frac{kq}{r - a \cos(\theta)} + \frac{kq}{r} - \frac{kq}{r + a \cos(\theta)}
\]
### Step 8: Final Result
After simplification, we find:
\[
V_P = \frac{2kq}{r} + \frac{2kqa \cos(\theta)}{r^2}
\]
Thus, the final expression for the electric potential at point P is:
\[
V_P = \frac{2kq}{r} + \frac{2kqa \cos(\theta)}{r^2}
\]
