Problem:

The finite sheet 0 < x="" />< 1,="" 0="" />< y="" />< 1="" on="" the="" z="0" plane="" has="" a="" charge="" density="" ρs="xy(x2"  ="" y2=""  ="" 25)3/2="" nc/m2="" .="" find="" the="" electric="" field="" at="" (0,="" 0,="" 5).="" explain="" in="" />

Solution:

Step 1: Identify the Symmetry

The problem has symmetry with respect to the z-axis. Therefore, the electric field can have no z-component and it can only point in the x-y plane.


Step 2: Calculate the Electric Field

The electric field is given by the following equation:
where ρs is the surface charge density and r is the distance from the point of interest to the element of charge.

Step 3: Break the sheet into pieces

To calculate the electric field, we break the sheet into small pieces and integrate over the entire sheet.


Step 4: Integrate the Electric Field

To integrate the electric field, we need to find the distance r from the point of interest to the element of charge. We can use the Pythagorean theorem to find r:

Next, we find the electric field at (0, 0, 5) due to the entire sheet by integrating over the entire sheet:

Step 5: Simplify the equation

We can simplify the equation by using the substitution u = x2   y2   25:

Step 6: Evaluate the integral

We can evaluate the integral by using the substitution u = x2   y2   25 and du = 50xy(x2   y2   25)1/2 dx dy:

Step 7: Simplify the equation

We can simplify the equation by using the substitution v = u3/2 and dv = 3/2u1/2 du:

Step 8: Evaluate the integral

We can evaluate the integral by using the substitution v = u3/2 and dv = 3/2u1/2 du:

Step 9: Simplify the equation

We can simplify the equation by using the substitution w = 1/v and dw = -dv/v2:

Step 10: Evaluate the integral

We can evaluate the integral by using the substitution w = 1/v and dw = -dv/v2:

Step 11: Simplify the equation

We can simplify the equation by using the substitution t = 5u1/2:

Step 12: Evaluate the integral