An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected to a battery and charged until its plates carry charges Q and - Q, and then disconnected from the battery. If the separation between the plates is now doubled, the potential difference between the plates will

Will be doubled.

Explanation:

For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:

where e₀ is a constant, the electric permittivity.

Now we can isolate V, the potential difference between the plates as:

Now, notice that the separation between the plates is in the numerator.

Thus, if we double the distance we will get a new potential difference V', such that:

So, if we double the distance between the plates, the potential difference will also be doubled.

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

answera compound;