Question

A parallel plate capacitor has square plates
of side 5 cm and separated by a distance of
1 mm. (a) Calculate the capacitance of this
capacitor. (b) If a 10 V battery is connected
to the capacitor, what is the charge stored
in any one of the plates? (The value of
ε = 8.85 x 10-12 Nm C 2)
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Answer 1

Explanation:

(a) The capacitance of the capacitor is

= 221.2 ×10−13 F

C = 22 . 12 ×10−12 F = 22 .12 pF

(b) The charge stored in any one of the plates is Q = CV, Then

= 22 . 12 ×10−12 ×10 = 221.2 ×10−12 C = 221.2 pC w

Answer 2

Answer:

A square-shaped parallel plate capacitor with an area is $0.0025m^{2}$ and separated by a distance of 1 mm the capacitance is  $22*10^{-12}F$, the charge stored when a $10v$ battery connected is $220*10^{-12}C$

Explanation:

• The parallel plate capacitor is two parallel plates that are connected across a battery when the plates are charged an electric field will induce between them.

• The CAPACITANCE is the ability to store electric charge within the capacitor

• For a parallel plate capacitor with an area $A$ separated by a distance $d$ in free space, the capacitance is

                                  $C=\frac{\varepsilon _0\ A}{d}$

• The amount of charge $Q$ that is stored in a capacitor of capacitance $C$ when a battery of voltage $V$  connected

                                  $Q=CV$                    

from the question, we have

the side length of the square plate $l=5cm=0.05m$

area of square plate $A=l^{2}=0.0025m^{2}$

distance between two parallel plates $d=1mm=0.001m$

voltage connected is  $V=10v$

capacitance in free space $C=\frac{8.8*10^{-12} *0.0025}{0.001} =22*10^{-12} F$

the charge stored in a plate $Q=22*10^{-12}*10=220*10^{-12}C$