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Alekssandra [29.7K]

3 years ago

11

Two charged particles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged par

ticle, with charge q3=q, is placed on the x-axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3. Find the position of charge 3 when q = 2.00 nC . Assuming charge 1 is located at the origin of the x-axis and the positive x-axis points to the right, find the two possible values x3,r, and x3,l for the position of charge 3.

1 answer:

Murrr4er [49]3 years ago

7 0

Answer:

X₃₁ = 0.58 m and X₃₂ = -1.38 m

Explanation:

For this exercise we use Newton's second law where the force is the Coulomb force

F₁₃ - F₂₃ = 0

F₁₃ = F₂₃

Since all charges are of the same sign, forces are repulsive

F₁₃ = k q₁ q₃ / r₁₃²

F₂₃ = k q₂ q₃ / r₂₃²

Let's find the distances

r₁₃ = x₃- 0

r₂₃ = 2 –x₃

We substitute

k q q / x₃² = k 4q q / (2-x₃)²

q² (2 - x₃)² = 4 q² x₃²

4- 4x₃ + x₃² = 4 x₃²

5x₃² + 4 x₃ - 4 = 0

We solve the quadratic equation

x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5

x₃ = [-4 ± 9.80] 10

X₃₁ = 0.58 m

X₃₂ = -1.38 m

For this two distance it is given that the two forces are equal

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Answer:

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Explanation:

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(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

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(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

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(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

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$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

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Hence, This is the required solution.

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