Determine the magnitude of the electric field at corner a

Posted By Admin @ September 03, 2022

Question:

Calculate the magnitude of the electric field at one corner of a square 2.42 m on a side if the other three corners are occupied by 4.25×10^−6 C charges. Express your answer to three significant figures and include the appropriate units.

Answer:

Ans:

12500 N/C

Explanation:

Side of square,  a = 2.42 m

q = 4.25 x 10^-6 C

The formula for the electric field is given by

E = \frac{Kq}{r^2}

where, K be the constant = 9 x 10^9 Nm^2/c^2 and r be the distance between the two charges

According to the diagram

BD = \sqrt{2}\times a

where, a be the side of the square

So, Electric field at B due to charge at A

E_{A}=\frac{Kq}{a^2}

E_{A}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}

EA = 6531.32 N/C

Electric field at B due to charge at C

E_{C}=\frac{Kq}{a^2}

E_{C}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}

Ec = 6531.32 N/C

Electric field at B due to charge at D

E_{D}=\frac{Kq}{2a^2}

E_{D}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2\times 2.42^2}

ED = 3265.66 N/C

Now resolve the components along X axis and Y axis

Ex = EA + ED Cos 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

Ey = Ec + ED Sin 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

The resultant electric field at B is given by

E=\sqrt{E_{x}^{2}+E_{y}^{2}}

E=\sqrt{8840.5^{2}+8840.5^{2}}

E = 12500 N/C

Explanation:

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