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Veronika [31]
3 years ago
11

There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12

V until it is loaded completely. Then the capacitor is disconnected from the voltage source. After this the two plates of the capacitor are driven apart until their distance is 5 mm. Now a positive test charge of 1 nC is brought from the positively charged plate to the negatively charges plate. How large is the kinetic energy of the test charge? The test charge of 1 nC can be regarded to be so small that it does not influence the electric field between the two plates of the capacitor.
Physics
1 answer:
Licemer1 [7]3 years ago
3 0

The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

        C = \frac{Q}{\Delta V}

        C = ε₀ \frac{A}{d}

we solve for the charge (Q)

        \frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

         Q = \epsilon_o \  \frac{A \ \Delta V_1 }{d_1}

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

         Q = \epsilon_o \  \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

          \frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}

          ΔV₂ = \frac{d_2}{d_1 } \ \Delta V_1

The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

          Em₀ = U = q DV₂

          Em₀ = q  \frac{d_2}{d_1 } \ \Delta V_1

           

final point. Proof load on the right plate

         Em_f = K

energy is conserved

         Em₀ = em_f

         q  \frac{d_2}{d_1 } \ \Delta V_1 = K

   

we calculate

         K = 1 10⁻⁹  12  \frac{0.005}{0.003}  

         K = 20 10⁻⁹ J

In this exercise, as the conditions at two different points of separation give, the area of ​​the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

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As the can heats, the compressed gases will expand, causing the can to explode

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A proton and an electron are released from rest, with only the electrostatic force acting. Which of the following statements mus
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Their kinetic energy will increase but potential energy will decrease.

Explanation:

Given that

Initial velocities of electron and proton is zero.

We know that ,electron have negative charge and proton have positive charge it means that they will attract to each other.We know that opposite charge attract to each other and same charge repels to each other.

It means that the velocities of proton and electron will increase and that leads to increase in the kinetic energy of proton and electron.

We know that potential energy U

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3 years ago
A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th
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Answer:

The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

The capacitance is modified as:

                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

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The power produced is 712.5 W

Explanation:

First of all, we start by finding the work done by the forklift.

The work done on an object pushed in the same direction as the force is given by:

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F is the magnitude of the force

d is the displacement of the object

In this problem,

F = 475 N is the force applied to raise the box

d = 18 m is the displacement

Substituting,

W=(475)(18)=8550 J

Now we can calculate the power, which is the ratio between the work done and the time taken:

P=\frac{W}{t}

Here we have,

W = 8550 J

t = 12 s

Substituting,

P=\frac{8550}{12}=712.5 W

Learn more about power here:

brainly.com/question/7956557

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