fundamentals of ph In the figure below, particles 1 and 2 of charge q_1 = q_2 =

Question

fundamentals of ph In the figure below, particles 1 and 2 of charge q_1 = q_2 = +4.80 times 10^-19 C are on a y axis at distance d = 14.0 cm from the origin. Particle 3 of charge q_3 = +9.60 times 10^-19 C is moved gradually along the x axis from x = 0 to x = +5.0 m. At what values of x will the magnitude of the electrostatic force on the third particle from the other two particles be (a) minimum and (b) maximum? What are the (c) minimum and (d) maximum magnitudes? position of minimum force _______ m position of maximum force _______ m magnitude of minimum force _______ N magnitude of maximum force _______ N

Explanation / Answer

Force due to q1:

F1 = k*q1*q/(d^2 +x^2)

similarly, F2 = k*q1*q/(d^2 +x^2)

So, the net force, Fnet = (k*q1*q/(d^2 + x^2))*x/sqrt(d^2+x^2)

So, Fnet = k*q1*q*(x/(d^2 + x^2)^1.5)

For maximum / minimum, dF/dx = 0

So, k*q1*q* [ (d^2+x^2)^-1.5 + x*(-1.5)*(d^2+x^2)^0.5*2x ] = 0

So, [ (d^2+x^2)^-1.5 + x*(-1.5)*(d^2+x^2)^-2.5*2x ] = 0

So, 1/(d^2+x^2)^1.5 = 3*x^2/(d^2+x^2)^2.5 <------ plugging in d = 14 cm = 0.14 m

So, x = +- 0.099 m or x = infinity

So, position of maximum = 0.099 m

and position of minimum = 0 m

c)

minimum force = 0 N

d)

maximum force = 9*10^9*9.6*10^-19*4.8*10^-19*0.099/(0.14^2 + 0.099^2)

= 1.4*10^-26 N

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