Well, first of all, there's no such thing as "fully charged" for a capacitor.
A capacitor has a "maximum working voltage", because of mechanical
or chemical reasons, just like a car has a maximum safe speed. But
anywhere below that, cars and capacitors do their jobs just fine, without
any risk of failing.
So we have a capacitor that has some charge on it, and therefore some
voltage across it. From the list of choices above . . .
<span>-- Both plates have the same amount of charge.
Yes. And both plates have opposite TYPES of charge.
One plate is loaded with electrons and is negatively charged.
The other plate is missing electrons and is positively charged.
-- There is a potential difference between the plates.
Yes. That's the "voltage" mentioned earlier.
It's a measure of how badly the extra electrons want to jump
from the negative plate to the positive plate.
-- Electric potential energy is stored.
Yes. It's the energy that had to be put into the capacitor
to move electrons away from one plate and cram them
onto the other plate.
</span>
Answer:
A(many people think that no energy or matter exists outside the universe)
Explanation:
Answer:
236.3 x 
C
Explanation:
Given:
B(0)=1.60T and B(t)=-1.60T
No. of turns 'N' =100
cross-sectional area 'A'= 1.2 x m²
Resistance 'R'= 1.3Ω
According to Faraday's law, the induced emf is given by,
ℰ=-NdΦ/dt
The current given by resistance and induced emf as
I = ℰ/R
I= -NdΦ/dtR
By converting the current to differential form(the time derivative of charge), we get
= -NdΦ/dtR
dq= -N dΦ/R
The change in the flux dФ =Ф(t)-Ф(0)
therefore, dq = 
(Ф(0)-Ф(t))
Also, flux is equal to the magnetic field multiplied with the area of the coil
dq = NA(B(0)-B(t))/R
dq= (100)(1.2 x )(1.6+1.6)/1.3
dq= 236.3 x C
