An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.60 mm. If a 15.0 V potential difference is applied to these plates, calculate the following.

(a) the electric field between the plates
kV/m

(b) the capacitance
pF

(c) the charge on each plate
pC

Given Information:

Area of separation = A = 7.60 cm² = 0.000760 m²

distance = d = 1.60 mm = 0.0016 m

Potential difference = V = 15.0 V

Required Information:

(a) Electric field = E = ?

(b) Capacitance = C = ?

(b) Charge = Q = ?

(a) Electric field = 9.375 kV/m

(b) Capacitance = 4.205 pF

(b) Charge = 63.08 pC

Explanation:

(a) The electric field between the plates  kV/m

We know that electric field is given by

E = V/d

E = 15.0/0.0016

E = 9.375x10³ V/m

or

E = 9.375 kV/m

(b) the capacitance  pF

The capacitance is given by

C = ε*A/d

where ε = 8.854×10⁻¹² F/m

C = 8.854×10⁻¹²*0.000760/0.0016

C = 4.205x10⁻¹² F

or

C = 4.205 pF

(c) the charge on each plate  pC

The charge on each plate is given by

Q = ε*E*A

Q = 8.854×10⁻¹²*9.375x10³*0.000760

Q = 63.08×10⁻¹² C

or

Q = 63.08 pC

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.