Find the electric field at a point P on the perpendicular bisector of a uniformly charged rod. The length of 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 point P on the perpendicular bisector of a uniformly charged rod, we can follow these steps:

### Step 1: Define the Problem
We have a uniformly charged rod of length \( L \) with total charge \( Q \). We want to find the electric field at point \( P \), which is located at a distance \( a \) from the center of the rod along the perpendicular bisector.

### Step 2: Set Up the Coordinate System
Place the rod along the x-axis, with its center at the origin. Therefore, the endpoints of the rod will be at \( -\frac{L}{2} \) and \( \frac{L}{2} \). The point \( P \) is located at \( (0, a) \).

### Step 3: Consider a Small Element of Charge
Consider a small element of the rod, \( dx \), located at a distance \( x \) from the center. The charge \( dQ \) on this element can be expressed as:
\[ dQ = \lambda \, dx \]
where \( \lambda = \frac{Q}{L} \) is the linear charge density.

### Step 4: Calculate the Distance from the Charge Element to Point P
The distance \( r \) from the charge element \( dQ \) to the point \( P \) is given by:
\[ r = \sqrt{x^2 + a^2} \]

### Step 5: Calculate the Electric Field Due to the Charge Element
The electric field \( dE \) due to the charge element \( dQ \) at point \( P \) is given by Coulomb's law:
\[ dE = k \frac{dQ}{r^2} = k \frac{\lambda \, dx}{x^2 + a^2} \]
where \( k \) is Coulomb's constant.

### Step 6: Resolve the Electric Field into Components
The electric field \( dE \) has both vertical and horizontal components. The horizontal components from symmetrical charge elements will cancel out, while the vertical components will add up. The vertical component is:
\[ dE_y = dE \cdot \frac{a}{\sqrt{x^2 + a^2}} = k \frac{\lambda \, dx}{x^2 + a^2} \cdot \frac{a}{\sqrt{x^2 + a^2}} = k \frac{\lambda a \, dx}{(x^2 + a^2)^{3/2}} \]

### Step 7: Integrate to Find the Total Electric Field
Now, integrate \( dE_y \) from \( -\frac{L}{2} \) to \( \frac{L}{2} \):
\[ E_y = \int_{-\frac{L}{2}}^{\frac{L}{2}} k \frac{\lambda a \, dx}{(x^2 + a^2)^{3/2}} \]

### Step 8: Solve the Integral
The integral can be solved as follows:
\[ E_y = k \lambda a \int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{dx}{(x^2 + a^2)^{3/2}} \]
This integral evaluates to:
\[ E_y = k \lambda a \left[ \frac{x}{a^2 \sqrt{x^2 + a^2}} \right]_{-\frac{L}{2}}^{\frac{L}{2}} = k \lambda a \left( \frac{\frac{L}{2}}{a^2 \sqrt{\left(\frac{L}{2}\right)^2 + a^2}} - \frac{-\frac{L}{2}}{a^2 \sqrt{\left(-\frac{L}{2}\right)^2 + a^2}} \right) \]
Simplifying this gives:
\[ E_y = \frac{2k \lambda L}{2a^2 \sqrt{\frac{L^2}{4} + a^2}} = \frac{k \lambda L}{a^2 \sqrt{\frac{L^2}{4} + a^2}} \]

### Step 9: Substitute for Linear Charge Density
Since \( \lambda = \frac{Q}{L} \), we substitute this into the equation:
\[ E_y = \frac{k Q}{a^2 \sqrt{\frac{L^2}{4} + a^2}} \]

### Final Result
Thus, the electric field at point \( P \) is given by:
\[ E = \frac{2k Q}{a \sqrt{L^2 + 4a^2}} \]