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Dafna11 [192]
2 years ago
12

The electric field in a region of space has the components Ey = Ez = 0 and Ex = (4.00 N/C · m) x. Point A is on the y axis at y

= 3.80 m, and Point B is on the x axis at x = 4.40 m. What is the potential difference VB − VA?

Physics
1 answer:
igor_vitrenko [27]2 years ago
5 0

Answer:

   V_{b}-V_{a} = -38.72

Explanation:

Consider the axis diagram attached.

Given:

                                        Ey = Ez = 0

                                       Eₓ = - 4x N/C · m

Since electric field is in x direction, potential difference would be:

              V_{b} - V_{a} =-\int\limits^{4.40}_{0} {E_{x} } \, dx

Here we integrate between limits 0 and 4.40 which is distance between A and B along x-axis.  

              V_{b} - V_{a} = -4 \left[\begin{array}{ccc}\frac{x^{2} }{2} \end{array}\right]^{4.40}_{0}

                       V_{b}-V_{a} = -4 (9.68)\\V_{b}-V_{a} = -38.72

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A high power line carries a current of 1.0 kA. What is the strength of the magnetic field this line produces at the ground, 10 m
solmaris [256]

Answer:

The strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

Explanation:

From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:

           

B = \frac{\mu_{0}I}{2\pi r} (1)      

                                     

Where \mu_{0} is the permiability constant, I is the current and r is the distance from the wire.    

             

Notice that it is necessary to express the current, I, from kiloampere to ampere.

I = 1.0kA \cdot \frac{1000A}{1kA} ⇒ 1000A

Finally, equation 1 can be used:

B = \frac{(4\pi x10^{-7}T.m/A)(1000A)}{2\pi (10m)}    

           

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Hence, the strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

         

8 0
2 years ago
Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a
raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0
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On my science test, there is a bonus question that I want to get right. Why would it be a bad idea to skydive on the moon? Hint:
eduard

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However, as it accelerates downward, its speed increases, and so does the air resistance F. However, F is upward, opposite to the direction of motion, therefore it reduces the net acceleration of the object; at a certain point, the magnitude of the air resistance will become equal to the weight, so that

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Explanation:

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