A charge Q is fixed on the X-Y plane at point (0, a). Find the electric field strength component along the X-axis, at any point (x, 0).

Answer

To find the electric field strength component along the X-axis at a point (x, 0) due to a charge Q fixed at the point (0, a), we can follow these steps:

### Step 1: Identify the position of the charge and the point of interest
The charge Q is located at the point (0, a) on the XY-plane, and we need to find the electric field at the point (x, 0) on the X-axis.

### Step 2: Calculate the distance between the charge and the point
The distance \( r \) between the charge at (0, a) and the point (x, 0) can be calculated using the distance formula:
\[ r = \sqrt{(x - 0)^2 + (0 - a)^2} = \sqrt{x^2 + a^2} \]

### Step 3: Write the expression for the electric field
The electric field \( E \) due to a point charge is given by the formula:
\[ E = \frac{kQ}{r^2} \]
where \( k \) is Coulomb's constant.

### Step 4: Substitute the distance into the electric field formula
Substituting \( r \) into the electric field formula gives:
\[ E = \frac{kQ}{(\sqrt{x^2 + a^2})^2} = \frac{kQ}{x^2 + a^2} \]

### Step 5: Determine the components of the electric field
Since we are interested in the component of the electric field along the X-axis, we need to consider the direction of the electric field vector. The electric field vector points away from the charge if the charge is positive and towards the charge if it is negative.

The electric field vector \( \vec{E} \) can be resolved into components. The X-component \( E_x \) can be found using:
\[ E_x = E \cdot \frac{x}{\sqrt{x^2 + a^2}} \]
Substituting \( E \) from the previous step:
\[ E_x = \frac{kQ}{x^2 + a^2} \cdot \frac{x}{\sqrt{x^2 + a^2}} = \frac{kQx}{(x^2 + a^2)^{3/2}} \]

### Final Result
Thus, the electric field strength component along the X-axis at the point (x, 0) is:
\[ E_x = \frac{kQx}{(x^2 + a^2)^{3/2}} \]