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Vaselesa [24]
2 years ago
14

The plate area is doubled, and the plate separation is reduced to half its initial separation. What is the new charge on the neg

ative plate
Physics
1 answer:
Norma-Jean [14]2 years ago
5 0

Answer:

 Q = 4 Q₀

Explanation:

This is an exercise on capacitors, where the capacitance is

         C =  \epsilon_{o} \    \frac{A}{d}

if we apply the given conditions

         C = \epsilon_{o} \    \frac{2A}{0.5d}

         C = 4 \epsilon_{o} \    \frac{A}{d}

let's call the capacitance Co with the initial values

         C₀ = \epsilon_{o} \    \frac{A}{d}

         C = 4 C₀

The charge on each plate of a capacitor is

           Q = C ΔV

If the potential difference is maintained, the new charge is

       Q = 4 C₀ ΔV

       

let's call

        Q₀ = C₀ ΔV

we substitute

        Q = 4 Q₀

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Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc
lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

E =  \frac{V}{L}

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

J = \frac{E}{p}

Substitute (V/L)  for E in the above equation of current density.

J = \frac{V}{pL} ------(1)

Substitute iR for V in equation (1)

J = \frac{iR}{pL} ------(2)

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

p = \frac{RA}{L}

A = \frac{pL}{R}

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A  and λ is use for (m/L) in the eqn above

\lambda = d\frac{p}{\frac{R}{L} } ------(3)

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³  for d and 1.69 × 10⁸ Ω .m

\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1

c) Using the equation (2) current density for aluminum cable is,

J = \frac{iR}{pL}

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

\lambda = d\frac{p}{\frac{R}{L} }

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0
3 years ago
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a man of mass 50 kg on the top floor of skyscraper step into an elevator what is the max weight as the elevator moves downward​
shtirl [24]
514.5 N is the answer
6 0
2 years ago
Your cat "Ms." (mass 7.00 {\rm kg}) is trying to make it to the top of a frictionless ramp 2.00 {\rm m} long and inclined upward
lbvjy [14]

Answer:

Final velocity at the top of the ramp is 6.58m/s

Explanation

Check the attachment

4 0
3 years ago
What is precision?what is precision? ​
RUDIKE [14]

Answer:

According to Oxford Dictionaries "Precision" means "the quality, condition, or fact of being exact and accurate."

Explanation:

Hope this helps! :)

6 0
2 years ago
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How much power is required to light a lightbulb at 100V of voltage when the lightbulb has a resistance of 500 Ohms?
salantis [7]

Answer:

Power = 20 Watts

Explanation:

Given the following data;

Voltage = 100 V

Resistance = 500 Ohms

To find the power that is required to light a lightbulb;

Mathematically, power can be calculated using the formula;

Power = \frac {Voltage^{2}}{resistance}

Substituting into the formula, we have;

Power = \frac {100^{2}}{500}

Power = \frac {10000}{500}

Power = 20 Watts

3 0
3 years ago
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