Question

Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diameter copper wire stretched taut from the center of one plate to the center of the other.a) What is the maximum current in the wire?
b) What is the largest electric field in the wire?
c) Does the current increase with time, decrease with time, or remain steady? Explain.
d) What is the total amount of energy dissipated in the wire?

Answer 1

Answer:

(a).The maximum current in the wire is $4.217\times10^{5}\ A$.

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$.

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times(5\times10^{-2})^2$

$A=7.85\times10^{-3}\ m^2$

We need to calculate the capacitor

Using formula of capacitor

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}$

$C=6.94\times10^{-12}\ F$

We need to calculate the resistance of the wire

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}$

$R=4.27\times10^{-3}\ \Omega$

We need to calculate the voltage

Using formula of charge

$q=CV$

$V=\dfrac{q}{C}$

Put the value into the formula

$V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}$

$V=1.801\times10^{3}\ V$

(a). We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

$I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}$

$I=421779.85\ A$

$I=4.217\times10^{5}\ A$

(b). We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}$

$E=11.2\times10^{5}\ N/C$

The electric field in the wire is $11.2\times10^{5}\ N/C$.

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

$E=\dfrac{1}{2}CV^2$

Put the value into the formula

$E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2$

$E=1.126\times10^{-5}\ J$

The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Hence, (a).The maximum current in the wire is $4.217\times10^{5}\ A$.

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$.

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$