A parallel plate capacitor of capacitance C0 has plates of area A with separation d between them. When it is connected to a battery of voltage V0, it has charge of magnitude Q0 on its plates. It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3. After the dielectric is added, the magnitudes of the capacitance and the potential difference between the plates are

The capacity increases and potential decreases with the introduction of dielectric .

Explanation:

The capacity of parallel plate capacitor when air is in between

C₀ = ε₀A/d

and when dielectric is introduced

The capacity C = Kε₀A/d

where K is dielectric constant .

When dielectric is introduced between the parallel plate capacitor , its capacity increases .

If the capacity of capacitor is C₀ initially . It will be 3 C₀ , after introducing the dielectric .

Thus if the initial potential V₀ = , the potential with dielectric is =

Thus the potential will become become 1/3 of initial value .

and the capacitance becomes 3 times the initial value .

battery with hi vtge high val resist in series

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s