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Fittoniya [83]
3 years ago
6

Trey is testing a properly working resistor in a circuit and finds its value to be far lower than he expects what’s a likely err

or
A. The resistor is faulty

B. The circuit is energized

C. He didn’t cut a lead on the resistor

D. The resistor is installed backwards

Physics
1 answer:
irinina [24]3 years ago
5 0
D. The resistor is installed backwards
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A positively charged particle is in the center of a parallel plate capacitor that has charge +/- Q on it's plates. Suppose the d
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Answer:

the force remains constant if the charge does not change

Explanation:

In a capacitor the capacitance is given by

           C = ε₀ A / d

Where ε₀ is the permissiveness of emptiness, A is about the plates and d the distance between them.

The charge on the capacitor is given by the ratio

            Q = C ΔV

Let's apply these expressions to our problem, if the load remains constant

            C = Q / ΔV = ε₀ A / d

            ΔV / d = Q / ε₀ A

If the distance increases the capacitance should decrease, therefore if the charge is a constant the voltaje difference must increase

Now we can analyze the force on the test charge in the center of the capacitor

               ΔV = E d

               E= ΔV/d

               F = q E

              F = q ΔV / d

 Let's replace

          F = q Q /ε₀ A

From this expression we see that the force is constant since the voltage increase is compensated by increasing the distance, therefore the correct answer is that the force remains constant if the charge does not change

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