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1 year ago

A conducting sphere of radius 5. 0 cm carries a net charge of 7. 5 µc. what is the surface charge density on the sphere?

Physics

1 answer:

elena-s [515]1 year ago

3 0

The surface charge density on the sphere is 0.014C/ $m^{2}$ .

Given,

r=5.00cm=0.05m, q=7. 5 µc

We know surface charge density σ =q/V= $\frac{7.5*10^{-6} }{\frac{4}{3}\pi 0.05^{3} }$ =0.014C/ $m^{2}$ .

<h3>Surface charge density </h3>

It is always important to know how charge is moving in an electric field. These fields will also build up electric charges. Therefore, estimation of the surface charge density is essential for a number of applications. It is also necessary to calculate the surface charge density of an electric object using its volume and surface area. The amount of electric charge that has collected in a given field is measured by the surface charge density. Using the dimensions provided, it determines the amount of electric charge. Dimensions of the electric body could be in the form of length, area, or volume. In one, two, or three dimensions, according to electromagnetism, surface charge density is a measurement of the amount of electric charge present in a given volume of space.

A conducting sphere of radius 5. 0 cm carries a net charge of 7. 5 µc. what is the surface surface charge density on the sphere?

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By using the kinematics equation to determine the speed of the water in the bucket and applying the equation of continuity to estimate the diameter of the column, we have the following;

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$\mathbf{v_f =12.29 \ m/s}$

From the equation of continuity:

$\mathbf{A_iV_i = A_fV_f}$

$\mathbf{\pi r^2_iV_i = \pi r^2_fV_f}$

$\mathbf{ r^2_iV_i = r^2_fV_f}$

$\mathbf{ (\dfrac{10}{2})^2\times 2.0 = r_f^2 \times 12.29}$

$\mathbf{ 50 = 12.29 \times r_f^2}$

$\mathbf{ r_f= \sqrt{\dfrac{50}{12.29} }}$

$\mathbf{ V_f= 2.02 \ cm }$

Since diameter = 2r;

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Answer:

Check the explanation

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b) f = n v / 4 L

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where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

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so we have

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making the average velocity

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Another way to find the average velocity is to compute it directly via

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d. We already know

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so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

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Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

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