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:
706.68 N
Explanation:
By Hooke's law,
Using the values in the question,
When e = 0.4 m,
Explanation:
Relation between potential energy and charge is as follows.
U = qV
or,
=
= J
or, =
Therefore, we can conclude that change in the electrical potential energy is .
Answer: KE = 25 J
Explanation: You must use the formula
KE = 1/2 m v²
to solve this problem.
KE = 1/2 (10 Kg) (5 m/s)
KE = 1/2 (50 kgm/s)
KE = 25 J