Question

Two 11-cm-diameter metal plates 1.0 cm apart are charged to ±12.5 nC. They are suddenly connected together by a 0.234-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?

c)What is the total amount of energy dissipated in the wire?

Answer 1

Given that,

diameter = 11 cm; so radius = r= 11/2 = 5.5 cm = 0.055 m

distance = d = 1 cm = 0.01 m; Charge = Q = 12.5 nC = 12.5 x 10^-9 C

diameter of the wire = 0.234 mm; r' = 0.234/2 = 0.000117m

(a)we need to find the maximum current. Let it be I.

The given setup is similar to parallel plate capacitor, and we know that capacitance is given by:

C = o A / d

C = 8.85 x 10^-12 x pi x (0.055)² / 0.01 = 8.4 x 10^-12 F = 8.4 pF

We know that, Q = CV => V = Q/C

V = 12.5 x 10^-9 C / 8.4 x 10^-12 F = 1.5 x 10³ V = 1500 Volts

We know from Ohm's Law that, V = IR => I = V/R

Now we need to know the resistance in order to find the current.

We know that, R = L/ A

R = 1.7 x 10^-8 x 0.01 / 3.14 x (0.000117)² = 0.017 x 10^-8 / 4.3 x 10^-8 = 0.004 Ohm

So, I =  1500 / 0.004 = 3.75 x 10⁵ Ampere.

Hence, the maximum current = I = 3.75 x 10⁵ Ampere.

(b) Energy dissipated would be given by:

E = 1/2 C V² = 0.5 x 12.4 x 10^-12 C x (1500)2 = 1.4 x 10^-5 J

Hence, energy dissipated = E = 1.4 x 10^-5 J.

Energy dissipated in the wire will be equal to Power:

E = P = I2 R = (3.75 x 10⁵ )² x 0.004 = 5.62 x 10⁸ J

Hence, E = 5.62 x 10⁸ J