Answer:
Explanation:
We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.
1 m =100 cm
Surface area =S=
We have to find the charge Q on the positive plates of the capacitor.
V=Initial voltage between plates
d=Initial distance between plates
Initial Capacitance of capacitor
Capacitance of capacitor after moving plates
Potential difference between plates after moving
Hence, the charge on positive plate of capacitor=
Answer:
Explanation:
the light ray leaving a medium in contrast to the entering or incident ray.
There are different options here but all of them work by approximating and assuming.
i) that the boulder is above ground.
ii) that the bottom surface of the boulder is known.
iii) the shape of the boulder is taken into account.
The most accurate way is measuring it by displacement method but the boulder is immovable hence the volume can be calculated by measuring the boulder or a waterproof box to be built around the boulder and calculate the volume occupied by boulder.
All the above methods are estimating methods.
*Another way to find the density is through specific gravity.
S.G = <u>Density</u><u> </u><u>of</u><u> </u><u>object</u>
Density of water
If the material that makes the boulder is known that is if it's stone or a mineral then the specific gravity can be found.
If the boulder is purely rock then S.G lies between 3 - 3.5 and the density of water is known thus the density of the boulder can be found without moving the boulder.
This is what I think after correction and allthe best!
Answer:
E/4
Explanation:
The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
Where;
E is the electric field
σ is the surface charge density
ε₀ is the electric constant.
Formula to calculate σ is;
σ = Q/A
Where;
Q is the total charge of the sheet
A is the sheet's area.
We are told the elastic sheet is a square with a side length as d, thus ;
A = d²
So;
σ = Q/d²
Putting Q/d² for σ in the electric field equation to obtain;
E = Q/(2ε₀d²)
Now, we can see that E is inversely proportional to the square of d i.e.
E ∝ 1/d²
The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.
From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;
E_new = E/4
Answer:
I want to say your answer is D) Electric
