Posted by Brette on Wednesday, November 7, 2012 at 2:48pm.
A positive charge q1 = 2.70 uC on a frictionless horizontal surface is attached to a
spring of force constant k as in the figure. When a charge of q2 = –8.60 uC is placed 9.50 cm away from the
positive charge, the spring stretches by 5.00 mm, reducing the distance between the charges to d = 9.00 cm.
Find the value of k.
physics - Jennifer, Wednesday, November 7, 2012 at 3:54pm
-k*x = -Ke*q1*q2/r^2
where k is the spring constant, x is the distance stretched by the spring, Ke is Coulomb's constant, r is the distance between the charges.
r = 9.00 cm = 0.09 m
x= 5 mm = 0.005 m
q1 = 2.70 uC = 2.7 * 10^-6 C
q2 = -8.6 uC = -8.6 * 10^-6 C
Ke = 8.9 * 10^9
physics - Anonymous, Tuesday, January 29, 2013 at 12:41am
4.62 x 10^3
physics - Sarah, Saturday, February 8, 2014 at 6:00pm
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