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Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge. Find the magnitude of the electric field at the center of the square. (k=1/4πϵ0=8.99×109 N · m2/C2)

Answer :

Answer:

4.30 x 10⁵ N/C

Explanation:

Two positive +3.0-μC point charges at opposite corners of the square creates equal and opposite electric field at the center. hence the electric field by these two positive charges at opposite corners becomes zero.

[tex]a[/tex] = length of the square of side = 0.50 m

[tex]r[/tex] = distance of the center from each corner = [tex]\frac{a}{sqrt(2)} = \frac{0.50}{sqrt(2)} = 0.354 m[/tex]

Magnitude of net electric field at the center is given as

[tex]E = \frac{2kq}{r^{2}} \\E = \frac{2(8.99\times10^{9})(3\times10^{-6})}{(0.354)^{2}} \\E = 4.30\times10^{5} NC^{-1}[/tex]

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