Just need final answer no need for explanation In ngure three positively charged

Question

Just need final answer no need for explanation
In ngure three positively charged particles are fixed on axis. Particles B and Care so close to other that t the x axis. In Figure (o), particle B has distance from particle force on partide A due to particles B and cris 3.92 x 10 23 N in the negative direction of the negative direction of the r axis. What is been moved to the opposite A but is still at the same distance from it. The net force on A is now 2.92 x 10 N in side of ratio QB? (a) Numbero

Explanation / Answer

force on a due to B and C combinely

F1 = Fc+Fb

when B is moved in opposite direction force on A is

F2 = Fc - Fb

by adding these equation

Fc =( F1+F2)/2

by subtracting

Fb = (F1-F2)/2

by dividing we get

Fc / Fb =( F1+F2)/(F1-F2)

Qc/Qb = (3.92+0.292)/(3.92-0.292)

Qc/Qb = 1.161

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