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:
2.30 × 10⁻⁸ N if the two electrons are in a vacuum.
Explanation:
The Coulomb's Law gives the size of the electrostatic force between two charged objects:
,
where
For the two electrons:
The sign of is negative. In other words, the two electrons repel each other since the signs of their charges are the same.
Answer:
a.) a = 0 ms⁻²
b.) a = 9.58 ms⁻²
c.) a = 7.67 ms⁻²
Explanation:
a.)
Acceleration (a) is defined as the time rate of change of velocity
Given data
Final velocity = v₂ = 0 m/s
Initial velocity = v ₁ = 0 m/s
As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,
a = 0 ms⁻²
b.)
Given data
As the space shuttle start from rest, So initial velocity is zero
Initial velocity = v₁ = 0 ms⁻¹
Final velocity = v₂ = 4600 ms⁻¹
Time = t = 8 min = 480 s
By the definition of Acceleration (a)
a = 9.58 ms⁻²
c.)
Given data
As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero
Initial velocity = v₁ = 0 ms⁻¹
Final velocity = v₂ = 4600 ms⁻¹
Time = t = 10 min = 600 s
By the definition of Acceleration (a)
a = 7.67 ms⁻²