Answer:
Q = 1.095 x 10^-9 C
Let the force experienced by the top piece of tape be F
F = kQ²/r²
r = distance between the two pieces tape = 1.00cm = 1.00 x 10^ -2 m
1/4(pi)*Eo = k = 8.99 x 10^9 Nm²/C²
The electric force of repulsion between the two charges and the weight of the top piece of tape are equal so
F = kQ²/r² = mg
Where m is the mass of the top piece of tape and g is the acceleration due to gravity
On re-arranging the equation above,
Q² = mgr²/k
Q² = ((11.0 x 10^-6) x 9.8 x (1.00x10^-2)²)/(8.99 x 10^9)
Q = 1.095x10^-9 C
Explanation:
The charge Q on both pieces of tape are equal and both act with a force of repulsion on each other.
The force of repulsion between both tapes pushes the top piece of tape upwards. The weight of the top piece of tape acts vertically downward. Since the top tape is in a position of equilibrium, the two forces acting on the top piece of tape must be equal to each other. This assumption is backed up by newton's first law of motion which states that the summation of all forces acting on a body at rest must be equal to zero. That is
Fe (electric force) - Fg (gravitational force) = 0
Fe = Fg
kQ²/r² = mg
On substituting the respective values for all variables except Q and rearranging the equation Q = 1.09 x 10^-9
