A dielectric material is inserted between the charged plates of a parallel-plate capacitor. Do the following quantities increase
1 answer:
Answer:
1. Charge on plates (plates remain connected to battery) increases.
2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases.
3.Capacitance (plates isolated from battery before inserting dielectric) increases.
4. Voltage between plates (plates remain connected to battery) remains the same.
5. Charge on plates (plates isolated from battery before inserting dielectric) remains the same.
6. Capacitance (plates remain connected to battery) increases.
7. Electric potential energy (plates remain connected to battery) increases.
8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases.
Explanation:
When a dielectric material is inserted between the plates of a capacitor, the capacitance is increase by a factor K, the dielectric constant.
By the capacitance formula, the other factors change accordingly.
1. Charge on plates (plates remain connected to battery) increases, because charge and capacitance are directly proportional.
2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases, because potential is inversely proportional to capacitance, and potential energy is given by the following formula
3.Capacitance (plates isolated from battery before inserting dielectric) increases.
4. Voltage between plates (plates remain connected to battery) stays the same, because the voltage is applied by the battery.
5. Charge on plates (plates isolated from battery before inserting dielectric) remains constant. If the plates isolated from the battery, then the total charge is conserved.
6. Capacitance (plates remain connected to battery) increases when a dielectric is inserted.
7. Electric potential energy (plates remain connected to battery) increases, because when plates remain connected to battery the voltage remains the same. But the capacitance increases. Therefore, electric potential energy increases.
8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases, because voltage is inversely proportional to capacitance.
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A:
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Then the acceleration equation is:
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To get the position equation we integrate again over time:
Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:
Now let's find the equations for car B.
We know that Car B does not accelerate, then it has a constant velocity given by:
To get the position equation, we can integrate:
This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:
Now we can answer this:
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We can solve this for t.
This is a quadratic equation, the solutions are given by the Bhaskara's formula:
We only care for the positive solution, which is:
Car A reaches Car B after 11.48 seconds.
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Here we only need to evaluate the position equation for Car A in t = 11.48s:
3) What is the velocity of car B when the two cars meet?
Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s
4) What is the velocity of car A when the two cars meet?
Here we need to evaluate the velocity equation for Car A at t = 11.48s