Capacitance C of a parallel plate capacitor is C=Q/V, where Q is the charge on the plates and V is the potential difference between the plates. The potential of the capacitor is then: V=Q/C.
The capacitance C is given by the equation C=k*ε₀*(A/d) where k is the dielectric constant, ε₀ is the permittivity of free space, A is the surface of the plate and d is the distance between the plates.
For vacuum and air, k=1 so the capacitance is C=ε₀*(A/d).
But when we insert a dielectric between the plates like teflon which has k=2.1 at T=20°C the capacitance becomes:
C₁=k*ε₀*(A/d)=2.1*ε₀*(A/d)=2.1*C.
We can see that the capacitance increases by k=2.1 which is the dielectric constant of teflon.
So the voltage V becomes:
V₁=Q/C₁=Q/(2.1*C)=(1/2.1)*(Q/C)=(1/2.1)*V=0.48*V.
V₁=0.48*V.
Potential difference decreases by 0.48 because we inserted teflon between the plates.
Answer:
The correct answer is "4.26 m".
Explanation:
Given:
Wavelength,
or,
Distance,
or,
Distance between the 1st and 2nd dark fringes,
As we know,
⇒
or,
⇒
By substituting the values, we get
Answer:
The height in which the book is dropped from is 1.28 m.
Explanation:
Mass of the book, m = 1.5 kg
initial velocity of the book, u = 0
final velocity of the bok, v = 5 m/s
The height in which the book is dropped from is given by;
v² = u² + 2gh
v² = 0 + 2gh
v² = 2gh
h = v² / 2g
h = (5)² / (2 x 9.8)
h = 1.28 m
Therefore, the height in which the book is dropped from is 1.28 m.
Answer:
10 seconds
Explanation:
As it starts from rest, then u=0
and by III rd equation of motion: