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

Answer:

(a). The charge is $1.5045\times10^{-11}\ C$

(b). The initial stored energy is $4.5135\times10^{-11}\ J$

(c).  The final stored energy is $12.036\times10^{-11}\ J$

(d). The work required to separate the plates is $7.5225\times10^{-11}\ J$

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

$C_{1}=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C_{1}=\dfrac{8.85\times10^{-12}\times8.5\times10^{-4}}{3.00\times10^{-3}}$

$C_{1}=2.5075\times10^{-12}\ F$

We need to calculate the charge

Using formula of charge

$Q=CV$

Put the value into the formula

$Q=2.5075\times10^{-12}\times6.00$

$Q=1.5045\times10^{-11}\ C$

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

Using formula of initial energy

$E_{i}=\dfrac{1}{2}\times CV^2$

$E_{i}=\dfrac{1}{2}\times2.5075\times10^{-12}\times36$

$E_{i}=4.5135\times10^{-11}\ J$

(c). We need to calculate the capacitance

Using formula of capacitance

$C_{2}=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C_{2}=\dfrac{8.85\times10^{-12}\times8.5\times10^{-4}}{2\times8.00\times10^{-3}}$

$C_{2}=9.403\times10^{-13}\ F$

We need to calculate the final stored energy

Using formula of initial energy

$E_{f}=\dfrac{1}{2}\times \dfrac{Q^2}{C}$

$E_{f}=\dfrac{1}{2}\times\dfrac{(1.5045\times10^{-11})^2}{9.403\times10^{-13}}$

$E_{f}=12.036\times10^{-11}\ J$

(d). We need to calculate the work done

Using formula of work done

$W=E_{f}-E_{i}$

Put the value in the formula

$W=12.036\times10^{-11}-4.5135\times10^{-11}$

$W=7.5225\times10^{-11}\ J$

Hence, (a). The charge is $1.5045\times10^{-11}\ C$

(b). The initial stored energy is $4.5135\times10^{-11}\ J$

(c).  The final stored energy is $12.036\times10^{-11}\ J$

(d). The work required to separate the plates is $7.5225\times10^{-11}\ J$