All categories

3 years ago

The linear charge density on the inner conductor is and the linear charge density on the outer conductor is

Physics

1 answer:

Lubov Fominskaja [6]3 years ago

5 0

Complete Question

The complete question is shown on the first uploaded image (reference for Photobucket )

Answer:

The electric field is $E = -1.3 *10^{-4} \ N/C$

Explanation:

From the question we are told that

The linear charge density on the inner conductor is $\lambda _i = -26.8 nC/m = -26.8 *10^{-9} C/m$

The linear charge density on the outer conductor is

$\lambda_o = -60.0 nC/m = -60.0 *10^{-9} \ C/m$

The position of interest is r = 37.3 mm =0.0373 m

Now this position we are considering is within the outer conductor so the electric field at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )

Generally according to Gauss Law

$E (2 \pi r l) = \frac{ \lambda_i }{\epsilon_o}$

=> $E = \frac{\lambda _i }{2 \pi * \epsilon_o * r}$

substituting values

$E = \frac{ -26 *10^{-9} }{2 * 3.142 * 8.85 *10^{-12} * 0.0373}$

$E = -1.3 *10^{-4} \ N/C$

The negative sign tell us that the direction of the electric field is radially inwards

=> $|E| = 1.3 *10^{-4} \ N/C$

You might be interested in

A stone is dropped from the

ICE Princess25 [194]

Height=h=500m
Acceleration=g=10m/s^2
Initial velocity=u=0
Speed of sound=c=340m/s
TIME TAKEN BY STONE TO HIT WATER=t
Time taken by sound to hear back=T

Now

$\\ \sf\longmapsto h=ut+\dfrac{1}{2}gt^2$

$\\ \sf\longmapsto h=0t+\dfrac{1}{2}10t^2$

$\\ \sf\longmapsto 500=5t^2$

$\\ \sf\longmapsto t^2=100$

$\\ \sf\longmapsto t=10s$

Now

$\\ \sf\longmapsto h=cT$

$\\ \sf\longmapsto T=\dfrac{h}{c}$

$\\ \sf\longmapsto T=\dfrac{500}{340}$

$\\ \sf\longmapsto T=1.47\approx 1.5s$

Total time:-

$\\ \sf\longmapsto T_{net}=t+T=10+1.5=11.5s$

8 0

1 year ago

What is the volume of an object that has a density of 65g/cm3 and a mass of 130g.

lora16 [44]

Density ρ is mass m per unit volume v, or

ρ = m / v

Solving for v gives

v = m / ρ

So the given object has a volume of

v = (130 g) / (65 g/cm³) = 2 cm³

5 0

2 years ago

Question 6. B) and c)

maks197457 [2]

The formula for both is v(t) = v0 + a*t

b) v(8) = 0 + 6m/s^2 *8s = 48 m/s

now we know the beginning (2) and end speed (14), but not the time:

c) 14 = 2 + 1.5*t => t = (14-2)/1.5 = 8 seconds

4 0

3 years ago

the train accelerates from 30 km/h to 45 km/h in 15 secs. a. find its acceleration. b. distance it travels during this time ...

Sonbull [250]

Acceleration = v-u/t
= (45-30)/15
= 1 km/h

Distance = ut + 1/2 at^2
s = [30 x 15] + 1/2 x 1 x 15^2
= 562.5 km

Hope this helps

3 0

2 years ago

Read 2 more answers

A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hol

il63 [147K]

Answer:

v = 12.1 m/s

Explanation:

When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.
These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.
In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.
Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.
This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.
At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):

$F_{c} = N + m*g (1)$

Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:

$F_{c} = m*\frac{v^{2}}{r} (2)$

Since the normal force takes any value as needed to make (1) equal to (2), if the speed diminishes, it will be needed less force to keep the equality valid.
In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.
In this condition, from (1) and (2), we can find the minimum possible value of the speed that still keeps the motorcycle touching the surface, as follows:
$v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)$

6 0

3 years ago