The potential difference between the plates of a parallel plate capacitor is 4.00 V. If the plate separation for this capacitor is 6.00 cm, what is the intensity of the electric field between the plates of this capacitor?

66.67 V / m

Explanation:

The electric field intensity (E) between the plates of a capacitor is the ratio of the potential difference (V) between the plates to the distance (d) of separation between the plates. Electric field intensity can be measured in V/m.  It can be written as follows;

E = V / d             (i)

From the question;

V = 4.00V

d = 6.00cm = 0.06m

Substitute these values into the equation above. i.e equation (i) as follows;

E = 4.00 / 0.06

E = 66.67 V / m

Therefore, the electric field intensity between the plates of the capacitor is 66.67 V / m

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Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.