Respuesta :
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 [tex]10^{9}[/tex] N m²/C²
e = (9 x [tex]10^{9}[/tex])(250 x [tex]10^{-7}[/tex]) / (0.5)²
e = 9 x [tex]10^{5}[/tex] N/C
Charge diagonal:
e = kq/r²
e = [(9 x [tex]10^{9}[/tex])(250 x [tex]10^{-7}[/tex]) / (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 [tex]10^{9}[/tex])(250 x [tex]10^{-7}[/tex]) / (0.5)²
e = 9 x [tex]10^{5}[/tex] N/C
Charge diagonal:
e = kq/r²
e = [(9 x [tex]10^{9}[/tex])(250 x [tex]10^{-7}[/tex]) / (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]
The electric field at one corner of the square due to the charge is 1.73 x 10⁶ N/C.
The given parameters;
- side of the square, L = 50 cm = 0.5 m
- magnitude of the charge, q = 250 x 10⁻⁷ C
The diagonal length of the square is calculated as follows;
[tex]d^2 = 0.5^2 + 0.5^2\\\\d^2 = 0.5\\\\d = \sqrt{0.5} \\\\d = 0.707 \ m[/tex]
The electric field in x-direction due to side length of 50 cm is calculated as;
[tex]E = \frac{kq}{r^2} \\\\E_x_1 = \frac{9\times 10^9 \times 250\times 10^{-7}}{(0.5)^2} \\\\E_x_1 = 900,000 \ N/C[/tex]
The electric field in y-direction due to side length of 50 cm is calculated as;
[tex]E = \frac{kq}{r^2} \\\\E_y_1 = \frac{9\times 10^9 \times 250\times 10^{-7}}{(0.5)^2} \\\\E_y_1 = 900,000 \ N/C[/tex]
The electric field due to the diagonal length of the square is calculated as;
[tex]E = \frac{kq}{r^2} \\\\E = \frac{9\times 10^9 \times 250\times 10^{-7}}{(0.707)^2} \\\\E= 450,135.94 \ N/C[/tex]
The diagonal length cut the square at an angle of 45⁰;
the x-component of the field;
[tex]E_x_2 = E cos (45)\\\\E_x_2 = 450,135.94 \times cos(45)\\\\E_x_2 = 318,291.12 \ N/C[/tex]
the y-component of the field;
[tex]E_y_2 = E cos (45)\\\\E_y_2 = 450,135.94 \times cos(45)\\\\E_y_2 = 318,291.12 \ N/C[/tex]
The net electric field;
[tex]E_x_{net} = 900,000 \ N/C \ +\ 318,291.12\ N/C = 1,218,291.12 \ N/C\\\\E_y_{net} = 900,000 \ N/C \ + \ 318,291.12 \ N/C = 1,218,291.12 \ N/C[/tex]
The resultant of the electric field is calculated as;
[tex]E = \sqrt{E_x^2 + E_y^2} \\\\E = \sqrt{(1,218,291.12^2 \ + \ 1,218,291.12^2} \\\\E = 1.73 \times 10^6 \ N/C[/tex]
Thus, the electric field at one corner of the square due to the charge is 1.73 x 10⁶ N/C.
Learn more here:https://brainly.com/question/22933709
