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3 years ago

Find the electric flux through a cylindrical Guassian surface produced by a point charge at base.

Physics

1 answer:

Simora [160]3 years ago

7 0

Answer:

Electric flux will be $\Phi =\frac{q}{\epsilon _0}$

Explanation:

Let the point charge enclosed by Gaussian surface is q

We have to find the electric flux through this Gaussian surface

According to Gauss law electric flux through any closed surface is equal to $\frac{1}{\epsilon _0}$ times of charged enclosed by the surface.

So electric flux will be $\Phi =\frac{q}{\epsilon _0}$

Gauss law is only valid for closed surface it is not applied for open surface

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Answer:

Following are the solution to the given question:

Explanation:

Its strength from both charges is equivalent or identical. The power is equal. And it is passed down

$F=\frac{kq_1q_2}{r^2}$

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$|F_{12}| = |F_{21}|$

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3 years ago

An ideal parallel-plate capacitor consists of a set of two parallel plates of area Separated by a very small distance 푑. This ca

dolphi86 [110]

Answer:

doubled the initial value

Explanation:

Let the area of plates be A and the separation between them is d.

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U = Q^2/2C ...(1)

Now the battery is disconnected, it means the charge is constant.

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C' = C/2

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3 years ago

How do solutions differ from other mixtures?

zalisa [80]

The answer is D.
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3 years ago

Read 2 more answers

A spring with a mass on the end of it hangs in equilibrium a distance of 0.4200 m above the floor. The mass is pulled down a dis

den301095 [7]

Answer:

0.48 m

Explanation:

I'm assuming that this takes place in an ideal situation, where we neglect a host of factors such as friction, weight of the spring and others

If the mass is hanging from equilibrium at 0.42 m above the floor, from the question, and it is then pulled 0.06 m below that particular position. This pulling is a means of adding more energy into the spring, when it is released, the weight compresses the spring and equals its distance (i.e, 0.06 m) above the height.

0.42 m + 0.06 m = 0.48 m

At the highest point thus, the height is 0.48 m above the ground.

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3 years ago

A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh

daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is $B=2.958\times10^{-5}T$

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

$\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \ \overrightarrow B_s\perp \overrightarrow B_w$

Angle of net magnetic field from axial direction is given by:

$tan\ \theta=\frac{B_w}{B_s}$ ,

Field due to solenoid:

$B_s=\mu_onI_s, \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m$

Field due to wire:

$B_w=\frac{\mu_oI_w}{2\pi r}$

Therefore, r:

$tan\ \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm$

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

$B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T$

Hence, the magnetic field is $B=2.958\times10^{-5}T$

7 0

3 years ago