When the charges in the rod are in equilibrium, what is the magnitude EEE of the electric field within the rod

Answers

Answer 1

Answer:

E= vB

Explanation:

If and when the charge on a rod happens to be in equilibrium, then we say that the electric force in conjunction with the magnetic force that is acting on the charge on the rod are both equal and opposite in direction. Mathematically, we say

Fe = Fm, where

Fe = qE and

Fm = qvB

If we substitute and make them equal to one another, we have

qE = qvB, and finally, on simplifying further, we have

E = vB

Answer 2

Answer:

Final answer:

When charges in a rod are in equilibrium, the magnitude of the electric field within the rod is zero. This principle stems from Gauss's law, which states that excess charge would be located on the surface of the conductor only, leaving the electric field basically zero in an equilibrium state.

Explanation:

The question is asking about the magnitude EEE of the electric field within a rod when the charges in the rod are in equilibrium. According to the principles of Physics, especially using Gauss's law, a key aspect to remember is that the electric field inside a conductor at equilibrium is essentially zero. This is because any excess charge would be located on the surface of the conductor only.

The magnitude of the electric field E is determined by the relationship:

E = qenc / 0

However, because the charge within the conductor, qenc, is zero in an equilibrium state, it implies that the magnitude of the electric field within the rod, E, is also zero. Hence, when a rod is in equilibrium, the magnitude of the electric field within the rod is zero.

Learn more about Electric Field here:

brainly.com/question/8971780

#SPJ3

Related Questions

What is the speed of an airplane that travels 4500 miles in 6 hours?
How can scientific method solve real world problems examples
A soccer ball is kicked straight upwards with an initial vertical speed of 8.0\,\dfrac{\text m}{\text s}8.0 s m 8, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. We can ignore air resistance. How long does it take the ball to have a downwards speed of 4.0\,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction?
The production of heat by metabolic processes takes place throughout the volume of an animal, but loss of heat takes place only at the surface (i.e. the skin). Since heat loss must be balanced by heat production if an animal is to maintain a constant internal temperature, the relationship between surface area and volume is relevant for physiology. If the surface area of a cube is increased by a factor of 2, by what factor does the volume of the cube change? Give your answer to two significant figures. 1.59
A moon is in orbit around a planet. The moon's orbit has a semimajor axis of 4.3 times 10 Superscript 8 Baseline m and has an orbital period of 1.516 days. Use these data to estimate the mass of the planet.

Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s. How much time does the driver of the car measure for his trip between the poles?

Answers

Answer:

Observed time, t = 5.58 s

Explanation:

Given that,

Speed of light in a vacuum has the hypothetical value of, c = 18 m/s

Speed of car, v = 14 m/s along a straight road.

A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.

We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :

$T=\frac{t}{\sqrt{1-(v^2)/(c^2)} }$

t is observed time.

$t=T* \sqrt{1-(v^2)/(c^2)} \n\nt=8.89* \sqrt{1-(14^2)/(18^2)} \n\nt=5.58\ s$

So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.

Anatomy of a Wave worksheet can someone help me out with the answers????

Answers

Part 1: here are the answers in order
5
2
1
3
4

Make the following conversion.56.32 kL = _____ L

0.056320
0.56320
5,632
56,320

Answers

Answer:

The answer would be D 56,320

Explanation:

A net force of 25.0 N causes an object to accelerate at 4.00 m/s2. What is the mass of the object?

Answers

Answer: 7kg I think or 6

Explanation:

Bryan is riding his scateboard at 2.0 m/s. He accelerates at 0.5 m/s for 4.0 seconds. Calculate his final velocity.

Answers

Answer:

4 m/s^2

Explanation:

(0.5m/s * 4)+2 m/s

A fish swims 10 cm from the front wall of an aquarium that is 35cm wide. The front wall of the aquarium is glass with negligible thickness, but the back wall is a plane mirror. A person looks through the front wall and watches both the fish and its reflection in the mirror.Part A:What is the apparent distance from the front wall of the aquarium to the fish?Part B: What is the apparent distance from the front wall of the aquarium to the image of the fish in the mirror?

Answers

El problema es un caso generalmente tipico en optica concerniente a Apparent depth vs real depth

We see the objects closer than their real depth to the surface. We see objects only if the rays coming from them reaches our eyes.

The equation is given by,

$D_a = (D_r)/(\eta)$

Where,

$D_a =$ Apparenth depth

$D_r =$ Real depth

$\eta =$ Refractive index of the medium of object

For water $\eta$ is equal to 1.33

I attach an image of the theory that could help clarify the measurements.

We have,

$D_a = (D_r)/(\eta)$

$D_a = (10)/(1.333)$

$D_a = 7.5cm$

Therefore the apparent distance between the front wall of the aquarium to the fish is 7.5cm

B) The distance between fish and mirror is given by,

$d=35-10= 25$

So we have that real distance from the front wall of to image of fish is

$dr=25+35=60cm$

Applying our equation we have that,

$D_a = (D_r)/(\eta)$

$D_a = (60)/(1.333)$

$D_a = 45.1cm$

Therefore the apparent distance from the front wall of the aquarium to the image of the fish is 45.1cm

Final answer:

The apparent distance from the front of the aquarium to the fish is 10 cm, and the apparent distance from the front of the aquarium to the image of the fish in the mirror is 35 cm.

Explanation:

Part A: When observing the fish in the aquarium, the apparent distance from the front wall of the aquarium to the fish is simply the actual distance. This is because the observation is being made directly through the glass which has negligible thickness. Therefore, the apparent distance to the fish is 10 cm.

Part B: The image of the fish in the mirror will seem farther away than the fish itself. This is due to the fact that light reflects off the mirror and travels the distance of the aquarium twice. Hence, the total distance traveled by the light is the distance to the fish plus the distance from the fish to the mirror which is 35 cm - 10 cm = 25 cm. Thus, the apparent distance from the front wall of the aquarium to the image of the fish in the mirror is 10 cm (to the fish) + 25 cm (to the mirror) = 35 cm.

Learn more about Apparent Distance and Refraction here:

brainly.com/question/32255407

#SPJ3

Other Questions

A 0.060 ???????? tennis ball, moving with a speed of 5.28 m/???? , has a head-on collision with a 0.080 ???????? ball initially moving in the same direction at a speed of 3.00 m/ ???? . Assume that the collision is perfectly elastic. Determine the velocity (speed and direction) of both the balls after the collision.

A spring oscillator is designed with a mass of 0.231 kg. It operates while immersed in a damping fluid, selected so that the oscillation amplitude will decrease to 1.00% of its initial value in 9.43 s. Find the required damping constant for the system.

In a region where there is a uniform electric field that is upward and has magnitude 3.80x104 N/C a small object is projected upward with an initial speed of 2.32 m/s The object travels upward a distance of 5.98 cm in 0 200 s. What is the object's charge-to-mass ratio q/m (magnitude and sign)?Assume g 9.80 m/s and ignore air resistance E3? C/kg q/m

You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond to the standing-wave resonances of the shower enclosure. A shower enclosure is created by adding glass doors and tile walls to a standard bathtub, so the enclosure has the dimensions of a standard tub, 0.75 m wide and 1.5 m long. Standing sound waves can be set up along either axis of the enclosure. What are the lowest two frequencies that correspond to resonances on each axis of the shower? These frequencies will be especially amplified. Assume a sound speed of 343 m/s.