Physics, 13.02.2020 18:18, domo1812

Calculate the electric field (magnitude and direction) at the upper right corner of a square 1.22 m on a side if the other three corners are occupied by 2.40×10−6 C charges. Assume that the positive x-axis is directed to the right.

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answered: Dreambig85

Explanation:

It is mentioned that the magnitude of given charge is $2.40 \times 10^{-6} C$ and side of the square is 1.22 m.

Hence, we will calculate the electric field due to charge at point A as follows.

$E_{a} = k\frac{q}{r^{2}}$

= $\frac{9 \times 10^{9} \times 2.4 \times 10^{-6}}{(1.22)^{2}}$

= $14.512 \times 10^{3} N/C$ (acts along y-axis)

Electric field at D due to the charge at C is as follows.

$E_{c} = k\frac{q}{r^{2}}$

= $\frac{9 \times 10^{9} \times 2.4 \times 10^{-6}}{(1.22)^{2}}$

= $14.512 \times 10^{3} N/C$ (acts along y-axis)

Electric field at D due to the charge at B is as follows.

$E_{d} = k\frac{q}{r^{2}}$

= $\frac{9 \times 10^{9} \times 2.4 \times 10^{-6}}{(1.7253)^{2}}$

= $7.2564 \times 10^{3} N/C$

Now, we will resolve it into two components.

$E_{bx} = E_{b} Cos 45 = 5.1318 \times 10^{3} N/C$

$E_{by} = E_{b} Sin 45 = 5.1318 \times 10^{3} N/C$

Now, the resultant x component is as follows.

$E_{x} = (14.512 + 5.1318) \times 10^{3} = 19.6438 \times 10^{3} N/C$

Resultant y component is given as follows.

$E_{y} = (14.512 + 5.1318) \times 10^{3} = 19.6438 \times 10^{3} N/C$

Therefore, resultant electric field at point D is as follows.

$E = \sqrt{E^{2}_{x} + E^{2}_{y}}$

= $\sqrt{(19.6438 \times 10^{3})^{2} + (19.6438 \times 10^{3})^{2}}$

= $27.780 \times 10^{3} N/C$

And, the direction of electric field is

$\theta = tan^{-1} (\frac{E_{y}}{E_{x}})$

= $tan^{-1} (1)$

= $45^{o}$

This means that net electric field makes an angle of $135^{o}$ with positive x-axis in counter clockwise direction.

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answered: Guest

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