A circular capacitor has radius 2mm and a separation between the plates of 3fim

Question

A circular capacitor has radius 2mm and a separation between the plates of 3fim and it is chargeed by a steady current of 4A. Since the electric field is changing we know there is magnetic field in between the plates. The magnetic field strengthe is zero in the center and largest at the circumference of the circle. Show me how you find the radius where the magnetic field at half it's maximum strength. What is that radius in meters? You must show your work. This problem is your opportunity to show me the physics concepts you know, take advantage of it, and include units.

Explanation / Answer

We will solve this problem step by step

We see that this problem has a high symmetry so we will use Ampere equation

B .ds = o ( I + Id)

Id = odE/dt

Data

R = 2 mm= 2 10-3 m

d = 3 m = 3 10-6 m

I = 4 A

When a current to a discharged capacitor is applied, a load on each plate of the same magnitude and opposite sign appears. These charge densities increases with time, so the electric field during load also increases, which creates a displacement current within the capacitor, of course the current I is zero because there is no movement of electrons.

B .ds = oodE/dt

As we use a flow path between the plates circle comdensador, the field is this circle, so in angle between ds is zero Cos B 0 = 1, I mean the scalar product is reduced to the ordinary product.

The flow of electric field

E = E. dA

The electric field is perpendicular to the plates and the normal vector is also perpendicular to the area, so the angle between the two is zero Cos 0 = 1, here also the regular product scalar product is reduced

B ds = o o d/dt E dA

We solve each side

circle ds = 2 dr

B ds = B ds = B 2 dr = B 2 r

o o d/dtEdA = o o d/dt ( E dA)

To take the integral use E of the separation between the plates is much smaller than its radius so the variation of E is very small

dA = 2 r dr

ood/dt ( E dA) = ood/dt ( E r2)

We rewrote the equation Ampere

B 2 r = oo r2dE/dt

B = ½ o o r dE/dt

Let's relate the electric field with a displacement current loads per unit time

E = Q / oA

Replace

B = ½ o o r 1/o A dQ/dt

The area is around the condenser A = R2

B = ½ o r 1/R2 dQ/dt

I = dQ/dt

B = o /2 r/ R2 I

in this expression we magnetic field dependence of the radius r

Bmax r=R

Bmax = o/2 1/RI

B= Bmax/2

Bmax/2 = o/2 r2/ R2I

Substituting the value of Bmax

½ (o/2 1/RI) = o/2 r2/ R2I

½ 1/R =r2/R2

r2 = ½ R

R2 = 1 mm = 1 10-3 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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