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To calculate the electric field at point \( P \) due to an electric dipole on the \( y \)-axis, let's analyze the setup and apply the relevant equations.

### Problem Setup
- There is a dipole on the \( y \)-axis with charges:
  - \( +Q \) at \( y = +d \)
  - \( -Q \) at \( y = -d \)
- The point \( P \) where we want to calculate the electric field is on the \( x \)-axis, a distance \( d \) from the origin.

### Solution
1. **Determine Distance to Point P**:
   - The distance from each charge \( \pm Q \) to point \( P \) is \( r = \sqrt{d^2 + d^2} = \sqrt{2}d \).

2. **Calculate Electric Field Due to Each Charge**:
   - The electric field \( E \) due to a point charge \( Q \) at a distance \( r \) is:
     \[
     E = \frac{k |Q|}{r^2} = \frac{k |Q|}{(2d^2)}
     \]
   where \( k = \frac{1}{4 \pi \epsilon_0} \) is Coulomb's constant.

3. **Component Analysis**:
   - Since \( +Q \) and \( -Q \) are symmetrically placed around the origin, their \( y \)-components of the electric field at \( P \) will cancel out. We only need the \( x \)-components.
   - The \( x \)-component of each field \( E_x \) is given by:
     \[
     E_x = E \cos \theta = \frac{k |Q|}{2d^2} \cdot \frac{d}{\sqrt{2}d} = \frac{k |Q|}{2d^2} \cdot \frac{1}{\sqrt{2}} = \frac{k |Q|}{2\sqrt{2} d^2}
     \]

4. **Total Electric Field at Point P**:
   - Since both \( +Q \) and \( -Q \) contribute equally in the \( x \)-direction, the total electric field \( E_{\text{total}} \) at \( P \) is:
     \[
     E_{\text{total}} = 2 \cdot E_x = 2 \cdot \frac{k |Q|}{2\sqrt{2} d^2} = \frac{k |Q|}{\sqrt{2} d^2}
     \]

### Final Answer
The magnitude of the electric field at point \( P \) is:
\[
E_{\text{total}} = \frac{k |Q|}{\sqrt{2} d^2}
\]

Would you like further explanation on any part of this solution? 

### Follow-up Questions:
1. What would the electric field be if \( P \) were twice the distance on the \( x \)-axis?
2. How does the direction of the electric field change if \( P \) is located at \( x = -d \)?
3. What would happen to the electric field if the dipole charges were doubled?
4. How does the electric field vary along the \( y \)-axis?
5. How does this solution change if we consider the electric potential at point \( P \) instead of the electric field?

### Tip:
For electric dipoles, always consider symmetry to simplify component calculations, especially when fields in one direction cancel out.

An electric dipole is placed on the y-axis with +Q at y = +d and -Q at y = -d. Calculate the magnitude of the electric field at point P located at a distance d on the x-axis.

Learn to calculate the electric field due to a dipole at a specified point on the x-axis. This example problem involves understanding electric fields from point charges, vector components, and applying the superposition principle. Suitable for college-level physics students.

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