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frozen [14]
3 years ago
7

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
Physics
1 answer:
Nina [5.8K]3 years ago
6 0

Answer:

a) 4

b) \frac{3}{4}

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 \frac{1}{k} times the energy without the dielectric

thus,

Final energy = \frac{1}{k} × 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 (\frac{100-25}{100})=\frac{3}{4}

c) Now,

removing half way out the dielectric

thus,

k' = k/2

or

k' = 4/2 = 2

and no charge escape

thus,

12C = k'\times C\times V'

or

V' =  \frac{12}{2}

or

V' = 6 volts

therefore,

the voltmeter will read 6 volts

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Ulleksa [173]

Answer:

The charges are q₁  = 2 × 10⁻⁸ C and  q₂ = 3 × 10⁻⁸ C

Explanation:

Here is the complete question

Two identical tiny balls have charge q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 X 10-4 N. after the balls are touched together and then represented once again to 20cm, now the repulsive force is found to be 1.40 X 10-4 N. find the charges q1 and q2.

Solution

The force F = 1.35 × 10⁻⁴ N when the charges are separated a distance of r = 20 cm = 0.2 m is given by

F = kq₁q₂/r₁²

q₁q₂ = Fr₁²/k

q₁q₂ = 1.35 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.054/9 × 10⁻¹³ C² = 0.006 × 10⁻¹³ C² = 6 × 10⁻¹⁶ C²

q₁q₂ = 6 × 10⁻¹⁶ C² (1)

When the charges are brought together, the charge is now q = (q₁ + q₂)/2

The new repulsive force F = 1.406 × 10⁻⁴ N  at a distance of r₂ = 20 cm = 0.2 m is then

F₂ = kq²/r₂²

q² = F₂r₂²/k = 1.406 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.00625 × 10⁻¹³ C² = 6.25 × 10⁻¹⁶ C²

q² = 6.25 × 10⁻¹⁶ C²

q = √(6.25 × 10⁻¹⁶) C

q = 2.5 × 10⁻⁸ C

(q₁ + q₂)/2 =  2.5 × 10⁻⁸ C

(q₁ + q₂) = 2 × 2.5 × 10⁻⁸ C

q₁ + q₂ = 5 × 10⁻⁸ C (2)

q₁  = 5 × 10⁻⁸ C - q₂  (3)

Substituting equation (3) into (1), we have

(5 × 10⁻⁸ C - q₂)q₂ = 6 × 10⁻¹⁶ C²

Expanding the bracket, we have

(5 × 10⁻⁸ C)q₂ - q₂² = 6 × 10⁻¹⁶ C²

So, q₂² - (5 × 10⁻⁸ C)q₂ + 6 × 10⁻¹⁶ C² = 0

Using the quadratic formula to find q₂

q_{2} = \frac{-(-5 X 10^{-8} )+/- \sqrt{(-5 X 10^{-8} )^{2} - 4X1X6 X 10^{-16} } }{2X1}\\  = \frac{5 X 10^{-8} )+/- \sqrt{25 X 10^{-16}  - 24 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- \sqrt{1 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- 1 X 10^{-8} }{2}\\= \frac{5 X 10^{-8} + 1 X 10^{-8} }{2} or \frac{5 X 10^{-8}  - 1 X 10^{-8} }{2}\\= \frac{6 X 10^{-8} }{2} or \frac{4 X 10^{-8}}{2}\\= 3 X 10^{-8} C or 2 X 10^{-8} C

q₁  = 5 × 10⁻⁸ C - q₂

q₁  = 5 × 10⁻⁸ C - 3 × 10⁻⁸ C or 5 × 10⁻⁸ C - 2 × 10⁻⁸ C

q₁  = 2 × 10⁻⁸ C or 3 × 10⁻⁸ C

So the charges are q₁  = 2 × 10⁻⁸ C and  q₂ = 3 × 10⁻⁸ C

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Answer:

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