Answer:
q = 1.815 \times 10^{-8} C
Charge on one plate is positive in nature and on the other plate it is negative in nature.
Explanation:
E = 8.20 x 10^5 V/m, A = 25 cm^2, d = 22.45 mm
According to the Gauss's theorem in electrostatics
The electric field between the two plates




Charge, q = surface charge density x area

q = 1.815 \times 10^{-8} C
Answer:
V = -RC (dV/dt)
Solving the differential equation,
V(t) = V₀ e⁻ᵏᵗ
where k = RC
Explanation:
V(t) = I(t) × R
The Current through the capacitor is given as the time rate of change of charge on the capacitor.
I(t) = -dQ/dt
But, the charge on a capacitor is given as
Q = CV
(dQ/dt) = (d/dt) (CV)
Since C is constant,
(dQ/dt) = (CdV/dt)
V(t) = I(t) × R
V(t) = -(CdV/dt) × R
V = -RC (dV/dt)
(dV/dt) = -(RC/V)
(dV/V) = -RC dt
∫ (dV/V) = ∫ -RC dt
Let k = RC
∫ (dV/V) = ∫ -k dt
Integrating the the left hand side from V₀ (the initial voltage of the capacitor) to V (the voltage of the resistor at any time) and the right hand side from 0 to t.
In V - In V₀ = -kt
In(V/V₀) = - kt
(V/V₀) = e⁻ᵏᵗ
V = V₀ e⁻ᵏᵗ
V(t) = V₀ e⁻ᵏᵗ
Hope this Helps!!!
The Joule is the SI unit of energy.
<span>The advantage of using an electromagnet rather than a regular magnet is that an electromagnet is stronger and you can control the power of the magnet. </span>
Answer:
a) the total circle of "initial" is Epot./Eel. and the total circle of "final" is Ekin.
b)intia: 0.5 * 50 * 4^2 = 400 J
final: 0 J = 0.5 * 50 * 0^2
Explanation:
c) in the initial phase the spring is compressed and so has Epot but then the Energie changes to be totally Ekin (in final phase) and so the Epot becomes 0