Answer:
Explanation:
E=(σ/ε0)
As noted by Dirac the field is the same no matter how far you are from the sheet. When your charge covers a conducting plane, as in your case, the field is, D/eo ,(D is charge density). Because the field inside the conductor (no matter how thin) is zero. The only time the field is, D/2eo, is when you have just a sheet of charge, by itself, not on a conducting plane."
Answer:
(a): The magnitude of the electric force on the small sphere = 
(b): Shown below.
Explanation:
<u>Given:</u>
- m = mass of the small sphere.
- q = charge on the small sphere.
- L = length of the silk fiber.
= surface charge density of the large vertical insulating sheet.
<h2>
(a):</h2>
When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:
<em>where,</em>
is the electrical permittivity of the free space.
The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:
Thus, the magnitude of the electric force on the small sphere is given by
The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.
<h2>
(b):</h2>
When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.
According to the fig.,
<em>where,</em>
= electric force on the sphere, acting along left.
= weight of the sphere, acting vertically downwards.
<em />
g is the acceleration due to gravity.
Answer:
The acceleration of the cart is 1.0 m\s^2 in the negative direction.
Explanation:
Using the equation of motion:
Vf^2 = Vi^2 + 2*a*x
2*a*x = Vf^2 - Vi^2
a = (Vf^2 - Vi^2)/ 2*x
Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.
Let x = Xf -Xi
Where Xf is the final position of the cart and Xi the initial position of the cart.
x = 12.5 - 0
x = 12.5
The cart comes to a stop before changing direction
Vf = 0 m/s
a = (0^2 - 5^2)/ 2*12.5
a = - 1 m/s^2
The cart is decelerating
Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.
I think the correct answer from the choices listed above would be the last option. It is the chemicals in the core of the star that cannot be determined from the spectrum of a star. Spectrum shows the different classification of the stars depending on their spectral characteristics. It usually involves the light, the wavelength and the distance.
