Can you answer one of the questions

A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

4 0

3 years ago

As a ship comes into view over the horizon the top appears before the rest of the ship how does this demonstrate the earth is sp

valina [46]

Since its a sphere, the top is seen first because its the tallest part if the ship. If the earth was flat, the whole ship would be seen.

7 0

3 years ago

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

Andreas93 [3]

Answer:

$v=1.08\times 10^7\ m/s$

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, $\dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2$

Separation between the plates, $d=1.75\times 10^{-2}\ m$

An electron is released from rest, u = 0

Using equation of kinematics,

$v^2-u^2=2ad$ ..........(1)

Acceleration of the electron in electric field, $a=\dfrac{qE}{m}$ ............(2)

Electric field, $E=\dfrac{\sigma}{\epsilon_o}$ ............(3)

From equation (1), (2) and (3) :

$v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}$

v = 10840393.1799 m/s

or

$v=1.08\times 10^7\ m/s$

So, the electron is moving with a speed of $1.08\times 10^7\ m/s$ before it reaches the positive plate. Hence, this is the required solution.

3 0

3 years ago

A boy standing throws a penny horizontally at 7.25 m/s out of the window of his apparent buliding. If the window is 10.0 m above

Ksivusya [100]

Answer:

Explanation:

This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".

In the y-stuff category:

v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)

Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)

a = -9.8 m/s/s

t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)

In the "x-stuff" category:

v₀ = 7.25 m/s (this is given)

Δx = ???

a = 0 (acceleration in this dimension is ALWAYS 0)

t = (we will solve for this in the y-dimension and plug it in here).

In the y dimension:

Δx = v₀t + $\frac{1}{2}at^2$ and plugging in from the y-dimension info:

$-10.0=0t+\frac{1}{2}(-9.8)t^2$ which simplifies to

$-10.0=-4.9t^2$ so

$t=\sqrt{\frac{-10.0}{-4.9} }$ which, to 2 significant digits is

t = 1.4 seconds

Now we will do the same in the x-dimension, using t = 1.4:

Δx = v₀t + $\frac{1}{2}at^2$ and filling in the x-stuff:

Δx = $7.25(1.4)+\frac{1}{2}(0)(1.4)^2$ Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:

Δx = 7.25(1.4) + 0 so

Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).

6 0

3 years ago

Read 2 more answers

What is the role of the water cycle in maintaining freshwater levels in lake and rivers?

Anika [276]

Replenishes Freshwater resources and moderates extremes in climate :))

6 0

3 years ago