In case of parallel plate capacitor, if the place separation is doubled and plate area is halved, the capacitance becomes A)4 times B)one half C)one forth D)zero

Explanation:

  The capacitance of a capacitor is defined as the charge stored per unit potential change.

using the expression for the capacitance

where e=  electrostatic constant

A = the area of the plate

d = separation distance.

Given the problem statement

1.  If the place separation is doubled

2. plate area is halved.

 effecting the conditions in our expression for the capacitance we have

cross multiplying we have

dividing both sides by 4d we have

The capacitance is 1/4 the initial value

(a): The moving car was going at Vo= 14.29 m/s moving toward the police car.

(b) If the police car had been traveling toward the other car at 20 m/s would receive a frequency of f''= 1327.39 Hz.

Explanation:

V= 343 m/s

f= 1200 Hz

f'= 1250 Hz

Vo= ?

Vf= 0

f'= f* {(V - (Vo) / V}

clearing Vo:

Vo= 14.29 m/s  (a)

f''= f* {(V - (Vo) / V - 20 m/s}

f''= 1327.39 m/s (b)