Question

two point charges 2 micro coulomb and 3micro coulomb are placed at two corners of an equilateral triangle of side 20cm in free space.calculate the magnitude of resultant electric field at third corner of the particle .If an alpha placed at the third corner,what is the force acting on it

Answer 1

Answer:

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Answer 2

Answer:

Force acting on it is F=$3.13*10^{-13}$

Explanation:

Two point charges are kept at 2 corners of equilateral triangle as shown in the figure. A third charge alpha particle is kept at third corner.

charge on alpha particle is =$2*1.6*10^{-19}C$

Force due 2 micro coulomb charge on alpha particle is $F_1=\frac{k q_1q_3}{r^2} =\frac{9*10^9*2*10^{-6}*2*1.6*10^{-19}}{400*10^{-4}} =1.44*10^{-13}N$

Similarly force due to 3 micro coulomb is $F_2=\frac{k q_3q_2}{r^2} =\frac{9*10^9*3*10^{-6}*2*1.6*10^{-19}}{400*10^{-4}} =2.16*10^{-13}N$

The resultant of F1 and F2 , $F=\sqrt{F_1^{2}+F_2^2+2F_1F_2cos60}$ putting the values we get F=$3.13*10^{-13}$.