Question

Three point charges, two positive and one negative, each having a magnitude of 20 C are placed at the vertices of an equilateral triangle (30 cm on a side). What is the magnitude of the electrostatic force on the negative charge?

Answer 1

The resultant force on the positive charge  is mathematically given as

X=40N

What is the magnitude of the electrostatic force on the negative charge?

Question Parameters:

Three-point charges, two positive and one negative, each having a magnitude of 20

Generally, the -ve charge   is mathematically given as

$Q+=\sqrt{x^2+x^2+2x.xcos120}\\\\Q+=\sqrt{2x^2+2x*(1/2)}$

Q+=X

Therefore

$x=\frac{Kq1q2}{r2}\\\\x=\frac{9*10^9*20*10^{-6}*20*10^{-6}}{(30*10^-2)^2}$

X=40N

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