.Answer:
The value of the work done is .
Explanation:
When a charged particle having charge is moving through an electric field , the net force () on the charge is
and the work done () by the particle is
Given, .
Substitute the value of electric field in equation (1) and then substitute the result in equation (2).
Answer:
136 N
Explanation:
Pressure on large piston = pressure on small piston
F₁ / A₁ = F₂ / A₂
1700 N / 1.5 m² = F / 0.12 m²
F = 136 N
Answer:
U/U₀ = 2
(factor of 2 i.e U = 2U₀)
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected
Explanation:
Energy stored in a capacitor can be expressed as;
U = 0.5CV^2 = Q^2/2C
And
C = ε₀ A/d
Where
C = capacitance
V = potential difference
Q = charge
A = Area of plates
d = distance between plates
So
U = Q^2/2C = dQ^2/2ε₀ A
The initial energy of the capacitor at d = d₀ is
U₀ = Q^2/2C = d₀Q^2/2ε₀ A ....1
When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.
The final energy stored in the capacitor at d = 2d₀ is
U = 2d₀Q^2/2ε₀ A ...2
The factor U/U₀ can be derived by substituting equation 1 and 2
U/U₀ = (2d₀Q^2/2ε₀ A)/( d₀Q^2/2ε₀ A )
Simplifying we have;
U/U₀ = 2
U = 2U₀
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.
Structure of Alveoli provides more surface area for lungs, to exchange the gases & for process of respiration