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telo118 [61]
3 years ago
13

A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the p

lanes x = d and x = -d. The y- and z-dimensions of the slab are very large compared to d; treat them as essentially infinite. The slab has a uniform positive charge density rho. (a) Explain why the electric field due to the slab is zero at the center of the slab (x = 0). (b) Using Gauss’s law, find the electric field due to the slab (magnitude and direction) at all points in space.
Physics
1 answer:
Anika [276]3 years ago
7 0

Answer:

Electric field at (x, y, z):  E=\frac{\rho x}{\epsilon_o}

Explanation:

(a) The slab is of an insulating material and has a uniform charge distribution. We can visualize this as infinite number of point charges, distributed throughout the slab, equally spaced apart. So if we (hypothetically) start to calculate the electric field due to each charge at x = 0, we shall always find a charge at a mirrored position about the x = 0 plane (within x = -d and x = d) and hence will cancel out the electric field.

A simpler example would be an infinitely long wire of uniform charge distribution. Any point on the wire will have zero electric field has there are essentially equal number of charges on either side (the length of the wire being infinitely long)

(b) Let us take a cylinder as a Gaussian surface with base area A. We shall take advantage of the symmetry about x = 0 and shall position the cylinder perpendicular to the y-z plane with x = 0 being the mid-point. Now the electric flux will only flow out through the 2 bases of the cylinder. This is because the slab has infinite dimensions along y and z-axes (think of an infinite <em>sheet </em>of charge) and the electric field always starts out perpendicular to any surface of charges.  If the Electric field at some point on the base of the cylinder be E, then total outgoing flux = 2EA

\rho is the charge density, hence,  Q_{enclosed}=\rho\times volume=2\rho Al

where 2l is the length of the cylinder and l is the x-coordinate.

Therefore, using Gauss's law,

2EA=\frac{2\rho Al}{\epsilon_o}

or, E=\frac{\rho l}{\epsilon_o}

or, E=\frac{\rho x}{\epsilon_o}

where, \epsilon_o = permittivity of free space.

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The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest st
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9.934 m/s²

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Two objects carry initial charges that are q1 and q2, respectively, where |q2| &gt; |q1|. They are located 0.160 m apart and beh
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Answer:

\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges  \rm q_1 and \rm q_1, separated by a distance \rm r, is given by

\rm F = \dfrac{kq_1q_2}{r^2}.

where k is the Coulomb's constant.

Initially,

\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, \rm F_f = +1.30\ N.

The new charges on the two objects are

\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.

The new force is given by

\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\

Using (1),

\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0

\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0

\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.

Using (1),

When \rm q_1 = \pm 8.00\times 10^{-7}\ C,

\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.

When \rm q_1=\pm 4.6\times 10^{-6}\ C,

\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.

Since, \rm |q_2|>|q_1|

Therefore, \rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.

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Katarina [22]

True

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