Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) what is the electric
1 answer:
The expression for the magnitude of the electric field between two uniform conducting plates is
Here, V is potential difference between plates and d is separation between plates.
As the potential 6.00 cm from the zero volt plate (and 4.00 cm from the other) is 420 V.
Therefore,
Thus, the electric field strength between the plates is 7000 V/ m
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Answer:
q = 8.61 10⁻¹¹ m
charge does not depend on the distance between the two ships.
it is a very small charge value so it should be easy to create in each one
Explanation:
In this exercise we have two forces in balance: the electric force and the gravitational force
F_e -F_g = 0
F_e = F_g
Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.
Let's write Coulomb's law and gravitational attraction
In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.
k q² = G m²
q = m
we substitute
q = m
q = m
q = 0.861 10⁻¹⁰ m
q = 8.61 10⁻¹¹ m
This amount of charge does not depend on the distance between the two ships.
It is also proportional to the mass of the ships with the proportionality factor found.
Suppose the ships have a mass of m = 1000 kg, let's find the cargo
q = 8.61 10⁻¹¹ 10³
q = 8.61 10⁻⁸ C
this is a very small charge value so it should be easy to create in each one