How are the electric field lines around a positive charge affected when a second positive charge is near it? The field lines com
2 answers:
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Answer:
Explanation:
The capacitance of the parallel-plate capacitor is given by
where
ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity
k = 3.00 is the dielectric constant
is the area of the plates
d = 9.00 mm = 0.009 m is the separation between the plates
Substituting,
Now we can calculate the energy of the capacitor, given by:
where
C is the capacitance
V = 15.0 V is the potential difference
Substituting,
To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.
According to Newton's second law we have to
where,
m= mass
g = gravitational acceleration
For the balance to break, there must be a mass M located at the right end.
We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.
In this way, applying the static equilibrium equations, we have to sum up torques at point B,
Regarding the forces we have,
Re-arrange to find M,
Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg
