The intensity of an electric field depends only on the coordinates x and y as follows,
`E = (a(xhati +y hatj))/(x^(2)+y^(2))` where, a is constant and `hati` and `hatj` are the unit vectors of the x and y axes. Find the charges within a sphere of radius R with the centre at the origin.

Explanation

To find the charge within a sphere of radius R centered at the origin, given the electric field intensity \( E = \frac{a(x \hat{i} + y \hat{j})}{x^2 + y^2} \), we will follow these steps:

### Step 1: Understand the Electric Field
The electric field \( E \) is given in terms of coordinates \( x \) and \( y \). It is a vector field that depends on the position in the xy-plane.

### Step 2: Define the Surface Area Element
For a sphere of radius \( R \), we need to define a small area vector \( dS \) on the surface of the sphere. The area vector \( dS \) is directed radially outward and can be expressed as:
\[ d\vec{S} = dS \hat{n} \]
where \( \hat{n} \) is the unit normal vector to the surface of the sphere. In spherical coordinates, this can be expressed as:
\[ \hat{n} = \frac{x \hat{i} + y \hat{j} + z \hat{k}}{R} \]

### Step 3: Calculate the Electric Flux
The electric flux \( \Phi_E \) through the surface of the sphere is given by the integral of the electric field over the surface area:
\[ \Phi_E = \int_S \vec{E} \cdot d\vec{S} \]
Substituting the expressions for \( \vec{E} \) and \( d\vec{S} \):
\[ \Phi_E = \int_S \frac{a(x \hat{i} + y \hat{j})}{x^2 + y^2} \cdot dS \hat{n} \]
This simplifies to:
\[ \Phi_E = \int_S \frac{a(x \hat{i} + y \hat{j})}{x^2 + y^2} \cdot \left( \frac{x \hat{i} + y \hat{j} + z \hat{k}}{R} dS \right) \]

### Step 4: Simplify the Dot Product
The dot product \( \vec{E} \cdot \hat{n} \) is:
\[ \vec{E} \cdot \hat{n} = \frac{a}{R} \left( \frac{x^2 + y^2}{x^2 + y^2} \right) = \frac{a}{R} \]
Thus, the flux becomes:
\[ \Phi_E = \int_S \frac{a}{R} dS \]

### Step 5: Calculate the Total Surface Area of the Sphere
The total surface area \( A \) of the sphere is:
\[ A = 4\pi R^2 \]
Therefore, the total electric flux is:
\[ \Phi_E = \frac{a}{R} \cdot 4\pi R^2 = 4\pi a R \]

### Step 6: Use Gauss's Law to Find the Enclosed Charge
According to Gauss's Law, the electric flux through a closed surface is equal to the charge enclosed \( Q \) divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi_E = \frac{Q}{\epsilon_0} \]
Setting the two expressions for flux equal gives:
\[ 4\pi a R = \frac{Q}{\epsilon_0} \]
Thus, the charge enclosed \( Q \) is:
\[ Q = 4\pi a R \epsilon_0 \]

### Final Answer
The charge within a sphere of radius \( R \) centered at the origin is:
\[ Q = 4\pi a R \epsilon_0 \]