Find the electric field at a point P on the perpendicular bisector of a uniformly charged rod. The lengthof the rod is L, the charge on it is Q and the distance of P from the centre of the rod is a.

Explanation

To find the electric field at a point P on the perpendicular bisector of a uniformly charged rod, we can follow these steps:

### Step 1: Understand the Configuration
We have a uniformly charged rod of length \( L \) with a total charge \( Q \). The point \( P \) is located at a distance \( a \) from the center of the rod along the perpendicular bisector.

### Step 2: Define the Charge Distribution
The charge per unit length \( \lambda \) of the rod can be defined as:
\[ \lambda = \frac{Q}{L} \]
This means that for a small segment \( dx \) of the rod, the charge \( dQ \) can be expressed as:
\[ dQ = \lambda \, dx = \frac{Q}{L} \, dx \]

### Step 3: Set Up the Electric Field Contribution
Consider a small segment \( dx \) of the rod located at a distance \( x \) from the center. The distance \( r \) from this segment to point \( P \) is given by:
\[ r = \sqrt{a^2 + x^2} \]
The electric field \( dE \) due to this small charge \( dQ \) at point \( P \) is given by Coulomb's law:
\[ dE = k \frac{dQ}{r^2} = k \frac{dQ}{a^2 + x^2} \]
where \( k \) is Coulomb's constant.

### Step 4: Resolve the Electric Field into Components
Since we are interested in the vertical component of the electric field (the component along the line connecting the rod to point P), we need to consider the angle \( \theta \) that the line from the charge segment to point P makes with the horizontal. We can express the vertical component as:
\[ dE_y = dE \cos \theta \]
Using the relationship \( \cos \theta = \frac{a}{r} \), we have:
\[ dE_y = dE \cdot \frac{a}{r} = k \frac{dQ}{a^2 + x^2} \cdot \frac{a}{\sqrt{a^2 + x^2}} = k \frac{a \, dQ}{(a^2 + x^2)^{3/2}} \]

### Step 5: Integrate Over the Length of the Rod
To find the total electric field \( E_y \) at point P, we integrate \( dE_y \) from \( -\frac{L}{2} \) to \( \frac{L}{2} \):
\[ E_y = \int_{-\frac{L}{2}}^{\frac{L}{2}} k \frac{a \, dQ}{(a^2 + x^2)^{3/2}} \]
Substituting \( dQ = \frac{Q}{L} \, dx \):
\[ E_y = \int_{-\frac{L}{2}}^{\frac{L}{2}} k \frac{a \frac{Q}{L} \, dx}{(a^2 + x^2)^{3/2}} \]
This simplifies to:
\[ E_y = \frac{kQ a}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{dx}{(a^2 + x^2)^{3/2}} \]

### Step 6: Evaluate the Integral
The integral can be evaluated using the substitution \( x = a \tan \theta \):
\[ dx = a \sec^2 \theta \, d\theta \]
This transforms the limits of integration and allows us to solve the integral.

### Step 7: Final Expression for Electric Field
After evaluating the integral, we find the expression for the electric field \( E_y \) at point P:
\[ E_y = \frac{2kQ}{L} \cdot \frac{a}{\sqrt{4a^2 + L^2}} \]

### Final Answer
Thus, the electric field at point P on the perpendicular bisector of the uniformly charged rod is:
\[ E = \frac{2kQ}{L} \cdot \frac{a}{\sqrt{4a^2 + L^2}} \]