Answer:
The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Explanation:
Given that,
dielectric constant k = 5.5
the area of each plate, A = 0.034 m²
separating distance, d = 2.0 mm = 2 x 10⁻³ m
magnitude of the electric field = 200 kN/C
Capacitance of the capacitor is calculated as follows;
Maximum potential difference:
V = E x d
V = 200000 x 2 x 10⁻³ = 400 V
Maximum energy that can be stored in the capacitor:
E = ¹/₂CV²
E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²
E = 6.62 x 10⁻⁵ J
Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Answer:
Neither.
Explanation:
When an electron is released from rest, in an uniform electric field, it will accelerate moving in a direction opposite to the field (as the field has the direction that it would take a positive test charge, and the electron carries a negative charge).
It will move towards a point with a higher potential, so its kinetic energy will increase, while its potential energy will decrease:
⇒ ΔK + ΔU = 0 ⇒ ΔK = -ΔU = - (-e*ΔV)
As ΔV>0, we conclude that the electric potential energy decreases while the kinetic energy increases in the same proportion, in order to energy be conserved, in absence of non-conservative forces.
If you are given distance and a period of time, you can calculate
the speed. The velocity of an object is the rate of change of its position with
respect to a frame of reference, and is a function of time. Velocity is
equivalent to a specification of its speed and direction of motion (e.g. 60
km/h to the north).
When Venus put in a different battery with higher voltage ... no matter
what other components were in the circuit ... the voltage across the
light bulb, and the current through it, both had to increase, and the
light bulb had to shine brighter than before.
Answer:pelo o que eu sei é ..
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Explanation:
