a parallel plate capacitor has a capacity 80 x 10-6 when air is present between the plates. the volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. the capacitor is now connected to a battery of 30 v by wires the dielectric slab is then removed. then the charge that passes now through the wire is

The charge that will pass through the wire will be approx. 0.003C.

Explanation:

Two equations are relevant to answer this question:

(1) The charge (Q) over a capacitance (C), given a voltage (U):

and

(2) the capacitance of a parallel-plate capacitor:

with being the dielectric constant, A the plate area and d their distance.

We are given the capacitance when air is present (80 micro-Farad, or uF). The dielectric constant for air/vacuum is 8.84 pico-Farad per meter. Then someone sneaky slides a slab of dielectric medium with dielectric constant of 20 pF/m and connects a source of 30V. By inserting the slab, they increased the capacitance by a factor of 20/8.84=2.26, making it 2.26*80=181uF. By connecting it to the source 30V, they charged our capacitor to a charge

but, the sneaky person changes his mind and removes the slab, effectively changing C down to the original 80uF. There is now excess charge on the capacitor that wants to leave! The steady-state charge on the air-filled capacitor should be:

The charge that will pass through the wire will be the difference:

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

don't remove my answer im trying my best to answer them

explanation:

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.