Which compound is held together by ionic bonds?
2 answers:
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Answer:
a) 4.33 pC b) 5.44*10² N/C
Explanation:
a) The vertical deflecting plates of an oscilloscope form a parallel-plate capacitor.
The value of the capacitance, for a parallel-plate capacitor with air dielectric, can be found to be as follows, applying Gauss' law to the surface of one of the plates, and assuming a uniform surface charge density:
C = ε₀*A / d
where ε₀ = 8.85*10⁻¹² F/m, A = (0.03m)², and d = 0.046 m (we assume that the informed value of 4.6 m is a typo, as no oscilloscope exists with this separation between plates).
Replacing by these values, we find the equivalent capacitance of the plates, as follows:
By definition, the capacitance of any capacitor can be expressed as follows:
where Q= charge on any of the plates, and V= potential difference between them.
As we know C and V, we can find Q as follows:
Q = C*V = 0.173*10⁻¹² F * 25.0 V = 4.33*10⁻¹² C = 4.33 pC
b) We can find the electric field in several ways, but one very easy is applying Gauss' Law to a pillbox with a face outside one of the plates (paralllel to it) and the other inside the surface.
The total electric flux through the surface must be equal to the enclosed charge, divided by ε₀.
If we look to the flux crossin any face, we find that the only one that has a non-zero flux, is the one outside the surface.
As the electric crossing the boundary must be normal to the surface (in electrostatic conditions, no tangential field can exist on the surface) , and we assume that the surface charge density that creates it is constant across the surface, we can write the Gauss ' Law as follows:
E*A = Q / ε₀
where A = area of the plate = (.03m)² = 9*10⁻⁴ m², Q= charge on one of the plates = 4.33*10⁻¹² C (as we found in a)) and ε₀ = 8.85*10⁻¹² N/C.
Replacing by these values, and solving for E, we have:
⇒ E = 5.44*10² N/C
We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
Let's start with energy conservation law:
This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
Because both carts have the same mass we can cancel those out:
Now we have our system of equation that we have to solve:
Part A
We need to solve our system for
. We will solve second equation for 
and then plug that in the first equation.
Now we have to plug this in the first equation:
We will multiply the first equation with 2 and divide by m:
Now we plug in the second equation into first one:
We end up with quadratic equation that we have to solve, I won't solve it by hand.
Coefficients are:
Solutions are:
Part B
We do the same thing here, but we must express from momentum equation:
Now we plug this into our energy conservation equation:
Again we end up with quadratic equation. Coefficients are:
Solutions are:
Let's start with energy conservation law:
This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
Because both carts have the same mass we can cancel those out:
Now we have our system of equation that we have to solve:
Part A
We need to solve our system for
Now we have to plug this in the first equation:
We will multiply the first equation with 2 and divide by m:
Now we plug in the second equation into first one:
We end up with quadratic equation that we have to solve, I won't solve it by hand.
Coefficients are:
Solutions are:
Part B
We do the same thing here, but we must express
Now we plug this into our energy conservation equation:
Again we end up with quadratic equation. Coefficients are:
Solutions are: