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3 years ago

A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.

Physics

1 answer:

tatiyna3 years ago

5 0

Answer:

a) x = ⅔ d , b) the charge must be negative, c) Q

Explanation:

a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing

∑ F = 0

-F₁₂ + F₂₃ = 0

F₁₂ = F₂₃

let's replace the values

k Q Q / r₁₂² = k Q 4Q / r₂₃²

Q² / r₁₂² = 4 Q² / r₂₃²

suppose charge 3 is placed at point x

r₁₂ = x

r₂₃ = d-x

we substitute

1 / x² = 4 / (d-x) 2

1 / x = 2 / (d-x)

x = 2 (x-d)

x = 2x -2d

3x = 2d

x = ⅔ d

b) The sign of the charge must be negative, to have an attractive charge on the two initial charges

c) Q

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2 years ago

what are the three types of solenoids, which one is the most common, and which one is used in pinball machines?

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Answer:

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Explanation:

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2 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

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2 years ago

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weqwewe [10]

Answer:

"h" signifies Planck's constant

Explanation:

In the equation energy E = h X v

The "h" there signifies Planck's constant

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It was named after Max Planck who discovered this unique relationship between the energy of a light wave and its frequency.

Planck's constant, "h" is usually expressed in Joules second

Planck's constant = $6.62607015 \times 10^{-34} J.s$

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