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3 years ago

6

What would the force be if the separation between the two charges in the top window was adjusted to 8.19 ✕10-11 m? (The animatio

n will not adjust that far--you will have to calculate the answer).
q1 = q2 = 1.00 ✕ e

1 answer:

dalvyx [7]3 years ago

3 0

The electrostatic force between the two charges is $3.4\cdot 10^{-8}N$

Explanation:

The electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have the following data:

$q_1 = q_2 = 1.00e$ is the magnitude of the two charges, where

$e=1.6\cdot 10^{-19}C$ is the fundamental charge

$r=8.19\cdot 10^{-11}m$ is the separation between the two charges

Substutiting into the equation, we find the force:

$F=(8.99\cdot 10^9)\frac{(1.00\cdot 1.6\cdot 10^{-19})^2}{(8.19\cdot 10^{-11})^2}=3.4\cdot 10^{-8}N$

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

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Answer:

The correct statement is "The electric field is directed toward the electron and has a magnitude of $\rm \dfrac{ke}{r^2}$ ".

Explanation:

According to Coulomb's law, the magnitude of the electric field due to a static point charge q at a point r distance away from it is given by

$\rm E = \dfrac{k|q|}{r^2}.$

k is the Coulmob's constant.

The direction of the electric field along the line joining the charge and the point where electric field is to be found and it is directed from positive charge to negative charge.

Conventionally, we assume a positive test charge placed at the point where electric field is to be found, the test charge has very small charge such that its charge does not affect the electric field due to the given charge.

The charge on the electron = -e.

The electric field due to an electron is given by

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The direction of this electric field is from positive test charge, placed at the point where electric field is to be found, towards the electron along the line joining the two.

Thus, the correct statement is "The electric field is directed toward the electron and has a magnitude of $\rm \dfrac{ke}{r^2}$ ".

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3 years ago

A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points

mr Goodwill [35]

Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

Weight of the book:

$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

$\mu_{\rm k}$ , the coefficient of kinetic friction, times
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That is:

$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

$\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}$ .

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