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In a rectangular coordinate system a positive point charge q = 6.00 x 10^-9C is placed at the point x = +0.150 m, y = 0, and an

makvit [3.9K]

Answer:

a) 0N/C

b) 2660.7N/C

c) 543.7N/C

Explanation:

To find the electric field in all these cases you take into account the x and y component of the total Electric field for each point.

for P(0,0), only there is the x component of the field because the point is the parallel line that connects both charges:

$E_x=E_{1x}-E_{2x}\\\\E_x=k\frac{q}{r_1^2}-k\frac{q}{r_2^2}$ ( 1 )

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r1: distance to the first charge = 0.150m

r:2 distance to the second charge = 0.150m

Due to in this case the distance r1=r2: you obtain, by replacing in (1):

$E_x=0$

for P(0.300 , 0 ) you also have only the x component of E, and the electric field generated by each charge are directed toward right:

$\vec{E}=E_x\hat{i}$

$E_x=k\frac{q}{r_1^2}+k\frac{q}{r_2^2}$

r1 = 0.300m+0.150m = 0.450m

r2 = 0.300m-0.150m = 0.150m

By replacing r1 and r2 in ( 1 ) you obtain:

$E_{x}=kq(\frac{1}{r_1^2}+\frac{1}{r_2^2})=(8.98*10^9Nm^2/C^2)(6.00*10^{-9})(\frac{1}{(0.450m)^2}+\frac{1}{(0.150m)^2})\\\\E_x=2660.7N/C\\\\\vec{E}=2660.7N/C\ \hat{i}$

for P(0.150 , -0.40) you have both x and y components for E:

$\vec{E}=E_x\hat{i}+E_y\hat{j}\\\\E_x=k\frac{q}{r_1^2}cos\theta+0N/C\\\\E_y=-k\frac{q}{r_1^2}sin\theta-k\frac{q}{r_2^2}$

the second charge does not contribute for the x component of E.

To find r1 you use Pitagora's theorem:

$r_1=\sqrt{(0.150+0.150m)^2+(0.40m)^2}=0.500m$

r2 = 0.40m

the angle is obtain by using a simple trigonometric relation:

$tan\theta=\frac{0.40}{0.150}=2.66\\\\\theta=tan^{-1}(2.66)=69.44\°$

Then, by replacing the values of r1, r1, q, theta and k you obtain:

$E_x=(8.98*10^9Nm^2/C^2)\frac{(6.00*10^{-9}C)}{(0.500m)^2}cos69.44=75.68N/C\\\\E_y=-(8.98*10^9Nm^2/C^2)(6.00*10^{-9}C)(\frac{sin69.44}{(0.500m)^2}+\frac{1}{(0.40m)^2})\\\\E_y=538.42N/C\\\\\vec{E}=75.68N/C\hat{i}-538.42N/C\hat{j}\\\\|\vec{E}|=\sqrt{(E_x)^2+(E_y)^2}=543.7N/C\\\\\theta=tan^{-1}(\frac{538.42}{75.68})=278\°$

hence, the magnitude of E is 543N/C with an angle of 278° from the positive x axis.

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Which types of energy is thermal energy a form of

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Renewable energy or a energy that can be used again

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Which of the following statements about electromagnetic waves in free space are true? (There could be more than one correct choi

Vanyuwa [196]

Answer:

The electric field carries the same mount of energy as the magnetic field

The frequency of the magnetic field is the same as the frequency of the electric field.

Explanation:

Let's analyze each option in detail:

The electric and magnetic fields have equal amplitudes. --> FALSE. In fact, the amplitudes of the electric and magnetic field are related by the equation:

$E=cB$

where

E is the amplitude of the electric field

B is the amplitude of the magnetic field

c is the speed of light

Therefore, from the equation we see that the amplitude of the electric field is much larger than that of the magnetic field.

The electric field carries the same mount of energy as the magnetic field. --> TRUE.

The energy carried by the electric field is:

$u_E = \frac{1}{2}\epsilon_0 E^2$

where $\epsilon_0$ is the vacuum permittivity.

The energy carried by the magnetic field is:

$u_B = \frac{1}{2\mu_0}B^2$

where $\mu_0$ is the vacuum permeability.

Given the following relationship:

$c=\frac{1}{\sqrt{\epsilon_0 \mu_0}}$

We can write

$E=cB=\frac{1}{\sqrt{\epsilon_0 \mu_0}}B$

$u_E = \frac{1}{2}\epsilon_0 (\frac{1}{\sqrt{\epsilon_0 \mu_0}}B)^2=\frac{1}{2\mu_0}B^2$

which means $u_E=u_B$ .

The electric field carries more energy than the magnetic field. --> FALSE, because in disagreement with the calculations above.

The frequency of the magnetic field is the same as the frequency of the electric field. --> TRUE. In an electromagnetic waves, electric field and magnetic field oscillate at the same frequency.

The frequency of the electric field is higher than the frequency of the magnetic field. --> FALSE, because in disagreement with the previous statement.

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15 points!

VMariaS [17]

Answer:

$4.12\times 10^{-5}\ J$ .

Explanation:

Given that,

Capacitance, $C=1.4\times 10^{-7}\ F$

Charge stored in the capacitor, $Q=3.4\times 10^{-6}\ C$

We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :

$E=\dfrac{Q^2}{2C}$

Put all the values,

$E=\dfrac{(3.4\times 10^{-6})^2}{2\times 1.4\times 10^{-7}}\\\\=4.12\times 10^{-5}\ J$

So, the required electric potential eenergy is equal to $4.12\times 10^{-5}\ J$ .

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3 years ago