A tiny sphere with a charge of 9.2 C is attached to a spring. Consider this posi

Question

A tiny sphere with a charge of 9.2 C is attached to a spring. Consider this position to be the origin. ONE other tiny charged sphere with a charge of 7.6 C, is placed in the position 18.2 cm directly vertically below and 13.3 cm to the right of the 9.2 C charge. The spring stretches 3.4 cm from its previous equilibrium position toward the 7.6 C.

Draw a diagram showing the position of the two charges and all the dimensions listed in the problem statement. Draw a free-body diagram showing the forces acting on the 9.2 C charge. Determine the (x, y) coordinates of the 9.2 C charge after the spring was stretched.

Calculate the spring force acting on the 9.2 C. If you do not know a quantity, leave it as an unknown. (Hint: Hooke’s Law.)

Calculate the Coulomb force acting on the 9.2 C caused by the 7.6 C charge.

From what you have done above; calculate the force constant for the spring.

Explanation / Answer

IN force diagram there will be two forces, one spring force toward the 9.2uC from -7.6 uC

other ias coloumb : direction-> from 9.2uC to 7.6uC

force will be along the line joining q1 and q2.

angle with x axis , @ = tan^-1(18.2 / 13.3) = 54.72 deg below x axis.

new position,

x = 0 + 3.4 cos(-54.72) = 1.96 cm

y = 0 + 3.4sin(-54.72) = - 2.78 cm ......Ans

for this position,

at equilibrium, Fnet= 0

spring force = couloumb force

distance between charges = sqrt[(18.2 - 2.78)^2 + (13.3 - 1.96)^2 ]
d = 19.14 cm = 0.1914 m

couloumb force = kq1q2 / d^2 = (9 x 10^9 x 7.6 x 10^-6 x 9.2 x 10^-6) / (0.1914^2)

     = 17.18 N   ........And (coloumb force and spring force)

spring force = kx = 17.18

k (0.034) = 17.18

k = 505.22 N /m   .....Ans (Spring constant)

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