A parallel plate capacitor has circular plates separated by 1 mm of air. Each pl

Question

A parallel plate capacitor has circular plates separated by 1 mm of air. Each plate has a radius of 2 cm. The capacitor is initially charged to a potential difference of 1000 V. At time equal to zero a switch is closed and the capacitor begins to discharge through a resistor of 0.20 ohms.

a) Find an expression for the strength of the magnetic field inside the capacitor as a function of r, the radial distance from the symmetry axis.
b) Draw a graph of the magnetic field versus time for r = 2 cm.
c) What is the initial value of the real current through the resistor immediately after the switch is closed?
d) What is the initial value of the displacement current through the capacitor immediately after the switch is closed?

Explanation / Answer

S=pi*2e-2^2=1.26e-3(m2). so that C=S*epsilon0/d=1.11e-11(F). a). we have that T=RC=2.22e-12(s). I=I0*e^(-t/T). where I0=1000/0.2=5000. Magnetic field inside the capacitor. I*r^2*muy0/r0^2=B*2*pi*r. so that B=I*r*muy0/2pi*r0^2 where I be current in the circuit r0 be 2cm and r be the radial distance. b)r=2cm. so B=I*muy0/2pi*r0. where I=I0*e^(-t/T). so B=I0*muy0*e^(-t/T)/2pi*r0. so that graph of B versus time is a e^(-x) graph. c) It is I0=1000/0.2=5000(ohms( d)the same as I0=5000ohms

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.