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Gelneren [198K]
3 years ago
12

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500

Physics
1 answer:
padilas [110]3 years ago
5 0

Answer:

I = 1.21x10^-5 A

Explanation:

You are missing the first part of the problem. This is an example, but it will give you the idea of how to solve yours with your data.

The first part is like this:

<em>A      4.0 cm  diameter parallel plate capacitor has a  0.44 m  m    gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s?</em>

Now with this, we can solve the problem.

In order to do this, we need to use the following expression:

q = CV (1)

Where:

C: Capacitance of a parellel capacitor (in Faraday)

q: charge of plate or capacitor (In coulombs)

V: voltage in Volts.

However, we need is the current, and we have data of potential difference, so, all we have to do is divide the expression between time so:

q/t = CV/t

And the current is q/t, thus:

I = C * V/t (2)

And finally, Capacitance C with two plates of area A separated by a distance d is:

C = Eo*A/d (3)

Where:

Eo = constant equals to 8.85x10^-12 F/m.

A = Area of the plate, in this case, πr²

d = gap of the capacitor.

Let's calculate first the Capacitance using equation (3):

C = 8.85x10^-12 * π * (0.04/2)² / 0.00046 = 2.42x10^-11 F

Now, it's time to use equation (2) and solve for I:

I = 2.42x10^-11 * 500,000

I = 1.21x10^-5 A

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A circular window of 30 cm diameter in a submarine can withstand a maximum force of 5.20 × 105 N. The maximum depth in a lake to
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Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

         F = 5.20 \times 10^{5} N

         g = 9.8 m/s

         radius = \frac{diameter}{2}

                    = \frac{30 cm}{2} = 15 cm = 0.15 m   (as 1 m = 100 cm)

Formula to calculate depth is as follows.

        F = \rho \times g \times h \times A

or,      h = \frac{F}{\rho \times g \times A}        

       h = \frac{5.2 \times 10^{5}}{1000 \times 9.8 \times (3.1416 \times (0.15 m^{2})}

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Thus, we can conclude that the maximum depth in a lake to which the submarine can go without damaging the window is closest 750 m.

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