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Lorico [155]
3 years ago
10

When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of 200 µC on each plate.

While the battery connection is maintained, a dielectric slab is inserted into, and fills, the region between the plates. This results in the accumulation of an additional charge of 250 µC on each plate. What is the dielectric constant of the dielectric slab?
Physics
1 answer:
Vika [28.1K]3 years ago
3 0

Answer:

The value is \epsilon =  2.25

Explanation:

From the question we are told that

The charge acquired on each plate is q =  200 \mu C  =  200 *10^{-6} \  C

The additional charge accumulated on each plate is q_a =  250 \mu C  =  250 *10^{-6} \ C

Generally the capacitance of the capacitor before a dielectric slab is inserted is

C_i =  \frac{q}{V}

=> C_i =  \frac{200 \mu C}{V}

Generally the total charge on each plate of the capacitor when the dielectric slab was inserted is

Q =  q + q_a

=> Q = ( 200 + 250 ) \mu C

=> Q = 450 \mu C

Generally the capacitance of the capacitor after a dielectric slab is inserted is mathematically represented as

C =  \frac{450 \mu C}{V}

Generally the dielectric constant is mathematically represented as

\epsilon =  \frac{C}{C_i}

=>      \epsilon =  \frac{450 \mu C}{200  \mu C}

=>       \epsilon =  2.25

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The total circuit current at the resonant frequency is 0.61 amps

What is a LC Circuit?

  • A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
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Q =15 = (wL)/R

wL = 30 ohms = Xl

R = 2 ohms

Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance

| Zs | = 30.07 <86.2° ohms

Xc = 1/(wC) = 30 ohms

The impedance of the LC circuit is found from:

Zp = (Zs)(-jXc)/( Zs -jXc)

Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°

I capacitor = 277/-j30 = j9.23 amps

I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps

I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps

Hence, the total circuit current at the resonant frequency is 0.61 amps

To learn more about LC Circuit from the given link

brainly.com/question/29383434

#SPJ4

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