Is this question this simple to answer? I feel like some of these questions are too easy....Makes me uneasy like I am missing something!
Two spherical point charges each carrying a charge of 40 mC are attached to the two ends of a spring of length 20 cm. If its spring constant is 120 Nm-1, what is the length of the spring when the charges are in equilibrium?
k = constant of proportionality
F = k ×(q1q2/r^2)
9x10^9 N×m2×C-2 × [(40μC^2)/0.2m^2]
F = 360N
k = spring constant = 120 N×m-1
F = kx
360N = 120 N×m-1 × x
x = 3m
When in equilibrium, the springs are stretched to 3 meters.
Two spherical point charges each carrying a charge of 40 mC are attached to the two ends of a spring of length 20 cm. If its spring constant is 120 Nm-1, what is the length of the spring when the charges are in equilibrium?
k = constant of proportionality
F = k ×(q1q2/r^2)
9x10^9 N×m2×C-2 × [(40μC^2)/0.2m^2]
F = 360N
k = spring constant = 120 N×m-1
F = kx
360N = 120 N×m-1 × x
x = 3m
When in equilibrium, the springs are stretched to 3 meters.