To find the electric field at the point (a, 0, 0) due to two point charges \( q \) and \( -q \) located at (0, 0, d) and (0, 0, -d) respectively, we can follow these steps:
### Step 1: Identify the Positions of the Charges and the Point of Interest
- The positive charge \( q \) is at position \( (0, 0, d) \).
- The negative charge \( -q \) is at position \( (0, 0, -d) \).
- The point where we want to calculate the electric field is \( (a, 0, 0) \).
### Step 2: Calculate the Distance from Each Charge to the Point
- The distance \( r_1 \) from the positive charge \( q \) to the point \( (a, 0, 0) \) is given by:
\[
r_1 = \sqrt{(a - 0)^2 + (0 - 0)^2 + (0 - d)^2} = \sqrt{a^2 + d^2}
\]
- The distance \( r_2 \) from the negative charge \( -q \) to the point \( (a, 0, 0) \) is:
\[
r_2 = \sqrt{(a - 0)^2 + (0 - 0)^2 + (0 + d)^2} = \sqrt{a^2 + d^2}
\]
### Step 3: Calculate the Electric Field Due to Each Charge
- The electric field \( E_1 \) due to the positive charge \( q \) at the point \( (a, 0, 0) \) is given by:
\[
E_1 = \frac{kq}{r_1^2} = \frac{kq}{a^2 + d^2}
\]
The direction of \( E_1 \) is away from the charge, which means it points in the positive x-direction.
- The electric field \( E_2 \) due to the negative charge \( -q \) at the point \( (a, 0, 0) \) is given by:
\[
E_2 = \frac{k(-q)}{r_2^2} = -\frac{kq}{a^2 + d^2}
\]
The direction of \( E_2 \) is towards the charge, which means it also points in the positive x-direction.
### Step 4: Determine the Net Electric Field
- Since both electric fields \( E_1 \) and \( E_2 \) are in the same direction (positive x-direction), we can add their magnitudes:
\[
E_{\text{net}} = E_1 + E_2 = \frac{kq}{a^2 + d^2} + \frac{kq}{a^2 + d^2} = \frac{2kq}{a^2 + d^2}
\]
### Step 5: Write the Final Expression for the Electric Field
- The net electric field at the point \( (a, 0, 0) \) is:
\[
E_{\text{net}} = \frac{2kq}{a^2 + d^2} \hat{i}
\]
where \( \hat{i} \) is the unit vector in the x-direction.
