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4 years ago

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Two point charges q and 4q are at x=0 and x=l, respectively, and free to move. a third charge is placed so that the entire three

-charge system is in static equilibrium. part a what is the magnitude of the third charge?

1 answer:

tatuchka [14]4 years ago

8 0

Explanation :

It is given that, q and 4q are placed at a distance of l.

Let x is the distance where third charge is placed so that the entire three charge system is in static equilibrium.

Equilibrium means the net force acting on the system of charges is equal to zero. Let Q is the third charge.

So, $\dfrac{kqQ}{x^2}=\dfrac{k(Q)(4q)}{(l-x)^2}$

On solving,

$x=\dfrac{l}{3}$

For magnitude :

$\dfrac{kqQ}{(l/3)^2}=\dfrac{kq4q}{l^2}$

using $x=\dfrac{l}{3}$

$Q=\dfrac{4q}{9}$

So, a charge of -4q/9 is at a distance of l/3 is placed. It is placed to the right of +q.

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an ideal gas system undergoes an adiabatic process in which it expands and does 20 J of work on its environment. how much energy

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Answer:

Zero.

Explanation:

An adiabatic process is one in which there is no exchange of heat energy. Therefore, in an adiabatic process, heat is neither added to the system not it is removed from the system.

The work done by the gas on the environment is 20 J. This energy is equal to the change in internal energy for an adiabatic process.

Therefore, for an ideal gas to undergo an adiabatic process in which it expands and does 20 J of work on its environment, the heat exchange is zero.

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3 years ago

At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q

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Answer:

$\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}$

$\displaystyle \theta =68^o$

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

$\displaystyle F=\frac{KQ_1Q_2}{d^2}$

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

$\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}$

$\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}$

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

$\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l$

The force F3 is

$\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}$

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

$\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle$

F2 is pointed upwards and its horizontal component is zero

$\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle$

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

$\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle$

$\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

Now we compute the total force

$\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

$\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle$

$\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}$

Now we compute the magnitude

$\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}$

The direction of the total force is given by

$\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558$

$\boxed{\displaystyle \theta =68^o}$

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The magnitude of the average force exerted on the ball by the wall is calculated below.

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Explanation:

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Answer:

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