Aparallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery. if the plates are pulled apart to a separation of 1.63 cm , suppose the battery remains connected while the plates are pulled apart. part a what is the capacitance?

at the beginning:

when the plates are pulled apart:

Explanation:

The capacitance of a parallel-plate capacitor is given by

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

is the area of the plates

is the separation between the plates at the beginning

Substituting into the formula, we find

Later, the plates are pulled apart to , so the capacitance becomes

object 2 because all the other ones are wrong

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