Answer:opposite
Explanation:for a capacitor to discharge (after charging) the polarities of the current and voltage have to be reversed
The question is incomplete. The complete question is :
A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .
Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:
1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.
2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.
Solution :
Given :
d = 10 mm
= 0.010 m
Then, Capacitance,
Now,
And
In parallel combination,
Then energy,
b). Now the charge on the is :
Now when the capacitor gets disconnected from battery and the is slowly of the way out of the is :
Without the dielectric,
Answer:
Explanation:
Angle of dip = 56° , magnetic field strength = 50μT
Vertical component = 50 x sin 56 = 41.45 μT
Horizontal component = 50 cos 56 = 27.96μT
New field is added in vertical downwards direction to increase the vertical component so as to increase the angle of dip . Let this field be B
total vertical field = B + 41.45
Horizontal component = 27.96
dip angle be θ
tanθ = vertical component / horizontal component
tan62 = B + 41.45 / 27.96
1.88 = B + 41.45 / 27.96
52.58 = B + 41.45
B = 11.13 μT
Since magnetic field has to be added , current should be clockwise when looked from above.
Answer:
The voltage across the capacitor decreases by a factor of 2.
Explanation:
As we know that capacitor is initially charges by battery connected across it
so we have
now capacitor is disconnected
so charge on the capacitor is conserved
Now a dielectric is inserted between the plates of the capacitor
So we will have
now the new voltage across the plates of capacitor is given as
so voltage between the plates of capacitor is decreased by factor of "k"
Answer:
Explanation:
After time "t" the angular position of A is given as
now we know that B start motion after time t1
so its angular position is also same as that of position of A after same time "t"
so we have
now since both positions are same