Please help with electronic field (algebra physic)
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Please help with electronic field (algebra physic)

[From: ] [author: ] [Date: 12-02-22] [Hit: ]
Er = square root of (3) x 9x10^9 x 20x 10^-6 / 4 = 7.8 x10^4 N/C.......
Positive point charges of +20uC are fixed at two of the vertices of an equilateral triangle with sides of 2.0m, located in vacuum. Determine the magnitude of the E-field at the third vertex

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the magnitude of electric field at the third vortex is Er = square root of (3) E

ie., Er = square root of (3) x 9x10^9 x 20x 10^-6 / 4 = 7.8 x10^4 N/C.
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