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A parallel plate capacitor is attached to a battery to create a potential difference of 12V The battery is then disconnected and a dielectric material is inserted to fill the gap with no loss of charge. A voltmeter then reads a difference of 3V 1. What is the dielectric constant of the material? 2. What fraction of energy was lost when the dielectric was inserted 3. If we pull the dielectric half way out with no read across the capacitor? charge escaping what woulda volt meter

Respuesta :

Answer:

a) 4

b) [tex]\frac{3}{4}[/tex]

c) 6 volts

Explanation:

Given:

Initial potential difference, V₁ = 12 V

Potential difference after inserting the dielectric, V₂ = 3 V

Now,

a) The charge is given as, q = CV

where, C is the capacitance

thus,

without dielectric, q = C × 12 = 12C     .........(1)

now, let the dielectric constant be 'k'

therefore,

q = kCV₂

or

12C = kC × 3          (q from 1)

or

k = 4

b) Now, the energy after inserting a dielectric with dielectric constant as k become the [tex]\frac{1}{k}[/tex] times the energy without the dielectric

thus,

Final energy = [tex]\frac{1}{k}[/tex] × Initial energy

or

Final energy = 0.25 × Initial energy

or

Final energy = 25% of the energy without dielectric

therefore,

hence, the energy loss fraction is [tex](\frac{100-25}{100})=\frac{3}{4}[/tex]

c) Now,

removing half way out the dielectric

thus,

k' = k/2

or

k' = 4/2 = 2

and no charge escape

thus,

12C = [tex]k'\times C\times V'[/tex]

or

V' =  [tex]\frac{12}{2}[/tex]

or

V' = 6 volts

therefore,

the voltmeter will read 6 volts

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