Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)

Help m please

The electric field at one corner of the square due to the charge is 1.73 x 10⁶ N/C.

The given parameters;

The diagonal length of the square is calculated as follows;

The electric field in x-direction due to side length of 50 cm is calculated as;

The electric field in y-direction due to side length of 50 cm is calculated as;

The electric field due to the diagonal length of the square is calculated as;

The diagonal length cut the square at an angle of 45⁰;

the x-component of the field;

the y-component of the field;

The net electric field;

The resultant of the electric field is calculated as;

Thus, the electric field at one corner of the square due to the charge is 1.73 x 10⁶ N/C.

Learn more here:link

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c²  =>  c = 0.707 m

Charge to the right:  In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x N m²/C²

e = (9 x )(250 x ) / (0.5)²

e = 9 x N/C

Charge diagonal:

e = kq/r²

e = [(9 x )(250 x ) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below:  For y direction

e = kq/r²

e = (9 x )(250 x ) / (0.5)²

e = 9 x N/C

Charge diagonal:

e = kq/r²

e = [(9 x )(250 x ) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

it is the second option.