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agasfer [191]
3 years ago
14

A cylindrical capacitor is being charged by an exponentially decreasing current. The inner cylinder of the capacitor is positive

and the outer cylinder is negative. What is the direction of the magnetic field between the two cylinders while the capacitor is charging
Physics
1 answer:
lisov135 [29]3 years ago
8 0

The direction of the magnetic field between the two cylinders while the capacitor is charging is counterclockwise.

According to Maxwell's law of induction, a changing electric field induces a magnetic field, B given by the equation

∫B.ds = μεdΦ/dt where ds = path length of magnetic field, μ = permeability of free space, ε = permittivity of free space and dΦ/dt = rate of change in electric flux. The electric field in the cylindrical capacitor is directed from the inner cylinder to the outer cylinder since the inner cylinder is positive and outer cylinder negative.

For an electric current which increases, the rate of change of electric flux is positive and this produces a magnetic field in the clockwise direction.

Since in this case, we have an exponentially decreasing current, dΦ/dt is negative. Thus, this will produce a magnetic field opposite to that of an increasing current. So, the direction of the magnetic field would be counter clockwise.

So, the direction of the magnetic field between the two cylinders while the capacitor is charging is counterclockwise.

Learn more about Maxwell's law of induction here:

brainly.com/question/4363096

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A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.
gavmur [86]

Answer:

(a) E = -1.02 \times 10^5~N/C

(b) E = -9.7 \times 10^4~N/C

Explanation:

(a) The electric field for a point charge is given by the following formula:

\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}

Hence, dE is now:

dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}

Therefore, vertical component of dE now becomes

dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z =  \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C

(b) We will have a similar approach, but a simpler integral.

dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta  = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi

E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

3 0
3 years ago
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