In the figure particles 1 and 2 of charge q1-q2 = +25.60 x 10-19 C are on a y ax

Question

In the figure particles 1 and 2 of charge q1-q2 = +25.60 x 10-19 C are on a y axis at distance d 19.5 cm from the origin. Particle 3 of charge q3-+20.80 x 10-19 C is moved gradually along the x axis from x = 0 to x = +6.86 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? 3 (a)Num (b) Number (c) Num (d) Numberl106.343E-29 0 Unit This answer has no units 2.9386 Units This answer has no units UnitsT N UnitsT N

Explanation / Answer

q = q1 = q2 = 25.6 E-19 C

q3 = 20.8 E-19 C

when x=0 btoh q1 and q2 exert equal force in opposite directions on q3, the net force is 0 and is minimum at x=0

as x increases the force on q3 from each of q1 and q2 is equal in magnitude but the directions are different along the line joining q1-q3 and q2-q3

The forces can be resolved along x and y

y-comp of each of the force is in opposite direction and cancle up the x-components act along +x and addup

force due to each source q1/q2

     = kqq3 /(x2 +d2)

resultant x-component from both sources

         = 2kqq3/(x2 +d2) * x/(x2 +d2)1/2

Fnet   = 2*9.0E+9 *25.6E-19 *20.8E-19 * x/(x2 +d2)3/2

           = 9.58E-26 * x/(x2 +d2)3/2

To find the max, we differentiate it wrt x and equal to 0

dF/dx   = 9.58E-26 * { -2x2 +d2 )/(x2 +d2)5/2 } =0

=> x = d/sqrt(2) = 13.79 cm

maximum force occurs at x= 13.79cm

Fmax = 9.58E-26 * 13.79E-2 /( (13.79E-2)2 + (19.5E-2)2 )3/2

          = 9.697E-25 N

Fmin =0

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