In the figure the three particles are fixed in place and hav

In the figure the three particles are fixed in place and have charges q_1 = q_2 = +2e and q_3 = +3e, Distance a = 3.64 mum. what is the magnitude of the net electric field at point p due to the particles?

Electric field due to charges Q₁ and Q₂ will cancel out each other since they have same magnitude and are in oppoisite directions.SO Net electric field is

E_net=E₃=KQ/r₃

since r₃=acos45 =3.64*10^-6*cos45

r₃=2.574*10^-6 m

E_net=(9*10⁹)(3*1.6*10^-19)/(2.574*10^-6)²

E_net=652.1 N/C

In the figure the three particles are fixed in place and have charges q_1 = q_2 = +2e and q_3 = +3e, Distance a = 3.64 mum. what is the magnitude of the net el

In the figure the three particles are fixed in place and hav

Solution

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