An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.5 cm, and the electric field within the capacitor has a magnitude of 3.9 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

Explanation:

When a q charge travels in an electric field , it acquires energy which is equal to V x q where V is potential difference and q is amount of charge

Here electric field E  = V / d where V  is potential difference between plate and d is plate separation

V = E x d

= 3.9 x 10⁶ x 1.5 x 10⁻²

= 5.85 x 10⁴ V

energy of electron = V x q  

= 5.85 x 10⁴ x 1.6 x 10⁻¹⁹ J

= 9.36 x 10⁻¹⁵ J  

Kinetic energy of electron = 9.36 x 10⁻¹⁵ J

the answer is b

explanation:

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.