Question

The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 3.00 mm, are charged by a 6.00 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.00 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates

Answer 1

The problem about a parallel-plate capacitor requires the application of electromagnetism principles to determine changes in potential difference, initial and final stored energy, and the work required to separate the plates, all based on the capacitor's geometry and the effect of plate separation on capacitance.

Step-by-step explanation:

To answer the student's question about the parallel-plate capacitor, we need to apply concepts from electromagnetism, specifically the relationships between charge, voltage, capacitance, and energy in capacitors. When the parallel plates of a capacitor are separated, the capacitance changes, but the charge remains the same since it is isolated after being disconnected from the battery. The potential difference between the plates changes as a result of the changing capacitance. The initial energy stored in the capacitor can be calculated using the formula U = 1/2 CV^2, and the final energy stored after increasing the plate separation can be calculated with the same formula but with the new capacitance value. The work done to separate the plates is equivalent to the change in stored energy, which can either be the work done by an external force or the work lost to the system.

Answer 2

(a). The charge is
`1.5045*10^(-11)\ C`

(b). The initial stored energy is
`4.5135*10^(-11)\ J`

(c). The final stored energy is
`12.036*10^(-11)\ J`

(d). The work required to separate the plates is
`7.5225*10^(-11)\ J`

Step-by-step explanation:

Given that,

Area = 8.50 cm²

Distance = 3.00 mm

Potential = 6.00 V

Distance without discharge = 8.00 mm

(a). We need to calculate the capacitance

Using formula of capacitance

`C_(1)=(\epsilon_(0)A)/(d)`

Put the value into the formula

`C_(1)=(8.85*10^(-12)*8.5*10^(-4))/(3.00*10^(-3))`

`C_(1)=2.5075*10^(-12)\ F`

We need to calculate the charge

Using formula of charge

`Q=CV`

Put the value into the formula

`Q=2.5075*10^(-12)*6.00`

`Q=1.5045*10^(-11)\ C`

(b). We need to calculate the initial stored energy

Using formula of initial energy

`E_(i)=(1)/(2)* CV^2`

`E_(i)=(1)/(2)*2.5075*10^(-12)*36`

`E_(i)=4.5135*10^(-11)\ J`

(c). We need to calculate the capacitance

Using formula of capacitance

`C_(2)=(\epsilon_(0)A)/(d)`

Put the value into the formula

`C_(2)=(8.85*10^(-12)*8.5*10^(-4))/(2*8.00*10^(-3))`

`C_(2)=9.403*10^(-13)\ F`

We need to calculate the final stored energy

Using formula of initial energy

`E_(f)=(1)/(2)* (Q^2)/(C)`

`E_(f)=(1)/(2)*((1.5045*10^(-11))^2)/(9.403*10^(-13))`

`E_(f)=12.036*10^(-11)\ J`

(d). We need to calculate the work done

Using formula of work done

`W=E_(f)-E_(i)`

Put the value in the formula

`W=12.036*10^(-11)-4.5135*10^(-11)`

`W=7.5225*10^(-11)\ J`

Hence, (a). The charge is
`1.5045*10^(-11)\ C`

(b). The initial stored energy is
`4.5135*10^(-11)\ J`

(c). The final stored energy is
`12.036*10^(-11)\ J`

(d). The work required to separate the plates is
`7.5225*10^(-11)\ J`