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Mrrafil [7]
3 years ago
15

A capacitor is connected to an ideal battery, and the capacitor is allowed to charge completely. The capacitor is then isolated

from the battery, and a material with a dielectric constant of 3 is slowly inserted between the plates of the capacitor. Which of the following correctly describes the process of inserting the material?
A) A force is required to push the material into the space between the plates, and the potential energy stored between the plates of the capacitor will remain the same.
B) A force is required to push the material into the space between the plates, and the potential energy stored between the plates of the capacitor will increase by a factor of 3
C) A force is required to push the material into the space between the plates, and the potential energy stored between the plates of the capacitor will increase by a factor of 9
D) The capacitor will pull the material into the space between the plates, and the potential energy stored between the plates of the capacitor will decrease by a factor of 3
E) The capacitor will pull the material into the space between the plates, and the potential energy stored between the plates of the capacitor will decrease by a factor of 9.

Physics
1 answer:
amm18123 years ago
8 0

Answer:

D) The capacitor will pull the material into the space between the plates, and the potential energy stored between the plates of the capacitor will decrease by a factor of 3

Explanation:

points to note

A) inserting a material with dielectric constant of 3 between the plates means that the capacitance C will increase by 3.

B) Since the battery is disconnected, the potential difference V between plate will not remain constant.

C) increasing the capacitance reduces the potential difference across the plate, and from C = Q/V it can be seen that the charges on the plate remains constant.

From the proof in the image below, the reduction in the potential energy of the capacitor is due to the energy used by the capacitor to pull the dielectric material into the space.

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Answer:

Approximately 3.99\times 10^{4}\; \rm J (assuming that the melting point of ice is 0\; \rm ^\circ C.)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}

The energy required comes in three parts:

  • Energy required to raise the temperature of that 0.100\; \rm kg of ice from (-10\; \rm ^\circ C) to 0\; \rm ^\circ C (the melting point of ice.)
  • Energy required to turn 0.100\; \rm kg of ice into water while temperature stayed constant.
  • Energy required to raise the temperature of that newly-formed 0.100\; \rm kg of water from 0\; \rm ^\circ C to 10\;\ rm ^\circ C.

The following equation gives the amount of energy Q required to raise the temperature of a sample of mass m and specific heat capacity c by \Delta T:

Q = c \cdot m \cdot \Delta T,

where

  • c is the specific heat capacity of the material,
  • m is the mass of the sample, and
  • \Delta T is the change in the temperature of this sample.

For the first part of energy input, c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}.

Similarly, for the third part of energy input, c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}.

The second part of energy input requires a different equation. The energy Q required to melt a sample of mass m and latent heat of fusion L_\text{f} is:

Q = m \cdot L_\text{f}.

Apply this equation to find the size of the second part of energy input:

\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}.

Find the sum of these three parts of energy:

\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}.

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