Question

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

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 μC charge. Determine the (x, y) coordinates of the 9.2 μC charge after the spring was stretched.

Calculate the spring force acting on the 9.2 μC. 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 μC caused by the − 7.6 μC charge.

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

Answer 1

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)