Question

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

Answer 1

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 = ?

Answer:

(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

Answer 2

Answer:

Explanation:

(a) For the calculation of the Electric field we use

$E=\frac{V}{d}=\frac{15.0V}{1.6*10^{-3}m} =9375\frac{N}{C}$

(b) The capacitance is calculate by using the expression

$C=\frac{\epsilon_{0}A}{d}=\frac{8.85*10^{-12}C^{2}/(Nm^{2})*(7.6*10^{-4}m^{2})}{1.6*10^{-3}m}=4.2*10^{-12}C$

(c) Finally, the charge on each plate is

$Q=CV=(4.2*10^{-12}C)(15V)=6.3*10^{-11}C$

I hope this is useful for you

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