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Elina [12.6K]
3 years ago
7

A vector points -43.0 units

Physics
1 answer:
Naddika [18.5K]3 years ago
8 0

Answer:

Explanation:

To find the direction of this vector we need o find the angle that has a tangent of the y-component over the x-component:

tan^{-1}(\frac{11.1}{-43.0})=-14.5 but since we are in Q2 we have to add 180 degrees to that angle giving us 165.5 degrees

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A 4.0 µC point charge and a 3.0 µC point charge are a distance L apart. Where should a third point charge be placed so that the
Rudik [331]

Answer:

Q must be placed at 0.53 L

Explanation:

Given  data:

q_1 = 4.0 μC , q_2 = 3.0μC

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The force by charge q_1 due to q is

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F1 = \frac{k q ( 4.0 μC )}{ x^2}                  ----1

The force by charge q_2 due to q is

F2 =  \frac{k q q_2}{(L-x)^2}

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we know that net electric force is equal to zero

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\frac{k q ( 4.0 μC )}{x^2}   =\frac{k q ( 3.0 μC )}{(l-x)^2}

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x = \sqrt{\frac{4}{3}*(L - x)

L-x = \frac{x}{1.15}

L = x + \frac{x}{1.15} = 1.86 x

x = 0.53 L

Q must be placed at 0.53 L

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