Question

A parallel-plate capacitor having plate area 25 cm2 and separation 1⋅00 mm is connected to a battery of 6⋅0 V. Calculate the charge flown through the battery. How much work has been done by the battery during the process?

Answer 1

Answer:

Given

Area, A=25 cm2 =25×10-4 m2

Voltage, V=6V

The separation between the plates is d= 1mm=1×10-3

Formula used:

When a capacitor is connected to a capacitor, the charge can be calculated

Where

Q is the charge of the capacitor

C is the capacitance of the capacitor

V is the Voltage or potential difference across the plates of the capacitor.

Capacitance C can be calculated by the formula

Where

C is the capacitance of the capacitor

D is the separation between the capacitor plates

A is the area of a circular plate capacitor

ε₀ is the permittivity of the free space,

When the capacitor is connected to a 6V battery, Charge flow through the battery is the same as the charge that can be withstand with the capacitor.

Substitute the value of C in 1)

Work is done by the battery

Where

W is the work done by the battery

Q is the charge on the plates of the capacitor

V is the voltage across the plates of the capacitor

Charge flows through the battery is

and work done by the battery is =8×10-10

Answer 2

The work has been done by the battery during the process is 8 $\times$ 10-10 J.

Explanation:

The capacitance of capacitor is given by  

C = €A/d

• Where € = dielectric constant of the capacitor
• A= Area of the plates
• d= distance between the plates

Area A = 25 cm2 = 25 $\times$ 10-4m

Distance d = 1 mm  = 1 $\times$ 10-3

Substituting the values =>

C = 8.85 $\times$ 10-12 $\times$ 25 $\times$ 10-4/1 $\times$ 10-3

After we get Capacitance C = 2.21$\times$10-11F

Given Voltage V = 6V

We know that Charge Q = C $\times$ V

Substituting the values =>

Q = 2.21 $\times$ 10-11 $\times$ 6

= 1.33 $\times$ 10-10 C

Hence work done is given by W = QV

W = 1.33 $\times$ 10-10 $\times$ 2.21 $\times$ 10-11

= 8 $\times$ 10-10

Hence, we can say that the work has been done by the battery during the process is 8 $\times$ 10-10 J.