Physics, 12.04.2020 05:46 carmen69

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)

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Answer from: maguilarz2005

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 mmagnitude of the charge, q = 250 x 10⁻⁷ C

The diagonal length of the square is calculated as follows;

$d^2 = 0.5^2 + 0.5^2\\\\d^2 = 0.5\\\\d = \sqrt{0.5} \\\\d = 0.707 \ m$

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

$E = \frac{kq}{r^2} \\\\E_x_1 = \frac{9\times 10^9 \times 250\times 10^{-7}}{(0.5)^2} \\\\E_x_1 = 900,000 \ N/C$

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

$E = \frac{kq}{r^2} \\\\E_y_1 = \frac{9\times 10^9 \times 250\times 10^{-7}}{(0.5)^2} \\\\E_y_1 = 900,000 \ N/C$

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

$E = \frac{kq}{r^2} \\\\E = \frac{9\times 10^9 \times 250\times 10^{-7}}{(0.707)^2} \\\\E= 450,135.94 \ N/C$

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

the x-component of the field;

$E_x_2 = E cos (45)\\\\E_x_2 = 450,135.94 \times cos(45)\\\\E_x_2 = 318,291.12 \ N/C$

the y-component of the field;

$E_y_2 = E cos (45)\\\\E_y_2 = 450,135.94 \times cos(45)\\\\E_y_2 = 318,291.12 \ N/C$

The net electric field;

$E_x_{net} = 900,000 \ N/C \ +\ 318,291.12\ N/C = 1,218,291.12 \ N/C\\\\E_y_{net} = 900,000 \ N/C \ + \ 318,291.12 \ N/C = 1,218,291.12 \ N/C$

The resultant of the electric field is calculated as;

$E = \sqrt{E_x^2 + E_y^2} \\\\E = \sqrt{(1,218,291.12^2 \ + \ 1,218,291.12^2} \\\\E = 1.73 \times 10^6 \ N/C$

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

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Answer from: sophiaa23

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 $10^{9}$ N m²/C²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (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 $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (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]

Answer from: Quest

pretty sure it's false

Answer from: Quest

i already answered it by 34/s

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