Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec
1 answer:
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
where
is the charge density
is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
where
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
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Answer:
1.8 cm
Explanation:
= mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg
= charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C
=Potential difference through which the ion is accelerated = 215 V
= Speed of the ion
Using conservation of energy
Kinetic energy gained by ion = Electric potential energy lost
= Radius of the path followed by ion
= Magnitude of magnetic field = 0.522 T
the magnetic force on the ion provides the necessary centripetal force, hence
Answer:
1.013 s
Explanation:
You can solve this problem using the equations for constant acceleration motion. The velocity at the bottom of the window can be found using this expression:
the gravity is negative as it opposes the movement.
Now, the time elapsed before the ball reappears is 2 times the time that it takes for the ball to go from the bottom of the window, reach maximum height, and reach again the bottom of the window, minus 2 times the time that it takes for the ball to travel from the top to the bottom of the window. The time that takes to the ball to reach maximum height, or in other words, to time that takes for the velocity of the ball to go from vo to 0m/s:
Then: