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alexandr1967 [171]
3 years ago
12

Four identical charges, Q, occupy the corners of a square with sides of length a. A fifth charge, q, can be placed at any desire

d location. Part A Find the location of the fifth charge.

Physics
1 answer:
Nat2105 [25]3 years ago
3 0

Answer:

  q = - ( 2*sqrt(2) + 1 )*Q / 4

Explanation:

Given:

- Side of each square L = a

- Charge q is placed among 4 other identical charges Q.

Find:

- Find the location and magnitude of the fifth charge q such that net Electric Force at its position is zero.

Solution:

- Compute distance r  from charge Q to q i.e center of all four charges Q:

                          r = 0.5*sqrt ( a^2 + a^2 )

                          r = a / 2sqrt(2)

- Compute the individual Electrostatic forces @ point A:

                          F_b = F_d = k*Q^2 / a^2

                          F_c = k*Q^2 / (a*sqrt(2))^2 = k*Q^2 / 2*a^2

                          F = k*Q*q / (a / 2sqrt(2))^2 = 2*k*Q^2 / a^2

- Use Electrostatic Equilibrium conditions:

                          F_b + F_c*cos(45) = F*cos(45)

                          F_d + F_c*sin(45) = F*sin(45)

- Plug in the values and equate:

                          { (Q/a)^2 + (Q^2 / 2*a^2*sqrt(2)) = sqrt(2)*Q*q / a^2

- Canceling all k's and a^2:

                          Q * ( 1 + (1 /2*sqrt(2)) ) =  sqrt(2)*q

                          q = - ( 2*sqrt(2) + 1 )*Q / 4

- Note: In attachment Q's and q's are interchange but the solution here provided is according to the question at hand.

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Answer:

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Explanation:

Using the Ideal Gases Law yoy have for pressure:

P_{1} = \frac{n_{1} R T_{1} }{V_{1} }

where:

P is the pressure, in Pa

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T is the temperature in Kelvin

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Given that the amount of material is constant in the process:

n_{1} = n_{2} = n

In an isobaric process the pressure is constant so:

P_{1} = P_{2}

\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }

\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }

V_{2} = \frac{T_{2} V_{1} }{T_{1} }

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Replacing on the ideal gases formula the pressure at this piont is:

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T = \frac{P V }{n R }

For the second process you have that T_{2} = T_{3}  So:

\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }

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Answer:

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From this equation; Φ = Q/ε₀

Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²

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Answer:

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Answer:

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6 0
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