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Anna11 [10]
2 years ago
8

S When an uncharged conducting sphere of radius a is placed at the origin of an x y z coordinate system that lies in an initiall

y uniform electric field →E = E₀ k^, the resulting electric potential is V(x, y, z)=V₀ for points inside the sphere andV(x, y, z)=V₀ - E₀z + E₀a³z / (x² + y² + z² )³/²for points outside the sphere, where V₀ is the (constant) electric potential on the conductor. Use this equation to determine the x, y , and z components of the resulting electric field (a) inside the sphere.
Physics
1 answer:
telo118 [61]2 years ago
5 0

The sphere has a constant potential. It is the electric field.

E = V_{0} = 0

In the sphere, then

E_{x} = 0,  E_{y}=0,   E_{z}=0

Outside the sphere, then

V = V_{0} - E_{0}z + \frac{E_{0}a^{3}z}{(x^{2} +y^{2} + z^{2})^{3/2}   }

The elements of the electric field include

E_{x} =\frac{3E_{0}a^{3}xy}{(x^{2} +y^{2} +z^{2})^{5/2}}\\E_{y} = \frac{3E_{0}a^{3}xz}{(x^{2} +y^{2}+z^{2})^{5/2}}

Which becomes,

=E_{0} (1-\frac{a^{3}}{x^{2} +y^{2}+z^{2})^{3/2}}+\frac{3a^{3}z^{2}}{(x^{2} +y^{2}+z^{2})^{5/2}})

<h3>In a consistent electric field, is force constant?</h3>

Similar to an ordinary object in the uniform gravitational field near the Earth's surface, a charged item in a uniform electric field experiences a constant force and consequently experiences a uniform acceleration. The vector cross product of p and E determines the torque's direction.

If the charge is positive, the force either moves in the same direction as E or in the opposite direction (if charge is negative).

A torque is experienced by an electric dipole (p) in an even electric field (E). The vector cross product of p and E determines the torque's direction.

To learn more about uniform electric field, visit

brainly.com/question/17426130

#SPJ4

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leonid [27]

The instant it was dropped, the ball had zero speed.

After falling for 1 second, its speed was 9.8 m/s straight down (gravity).

Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.

Falling for 1 second at an average speed of 4.9 m/s, is covered <em>4.9 meters</em>.

ANYTHING you drop does that, if air resistance doesn't hold it back.

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3 years ago
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A 150 g baseball is traveling horizontally at 50 m/s. If the ball takes 20 ms to stop once it is in contact with the catcher’s g
Sliva [168]
To solve for force, you need to get the product of mass and acceleration. 
F = ma

Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms

As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration. 
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a = change in velocity/time

To get the change in velocity, you get the difference between the initial velocity and final velocity:

a =  \frac{vf-vi}{t}

The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s. 

So we can put it into our formula now:

a = \frac{0m/s-50m/s}{20ms}

WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s. 

20ms x \frac{1s}{1000ms} = \frac{20s}{1000} = 0.02s

So your new time is 0.02s. Now we put this time into the formula:


a = \frac{0m/s-50m/s}{0.02s}
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As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop. 

So now we have our acceleration. Now using this, we can get our force. 

F= ma

Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is kg.m/ s^{2} but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first. 

150g =  \frac{1kg}{1,000g}  =  \frac{150kg}{1,000}  = 0.150kg

Our new mass is 0.150kg. 

To make things clearer, let us write down all our new values:

m = 0.150kg
a = -2,500 m/ s^{2}

Now that all our units match, we can put that into our formula:

F= ma
F= (0.150kg)(-2,500m/s^{2})
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The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign. 



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3 years ago
A certain car engine delivers enough force to create 630 N⋅m of torque when the engine is operating at 3200 revolutions per minu
jekas [21]

The appropriate expression for the calculation of power by relating the angular energy in a given time.

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P=\tau\omega

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Our values are given by

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Replacing,

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Physical property is defined as the property of a substance which becomes evident during physical change in which there is alteration in shape, size etc. No new substance gets formed during physical change.

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Chemical property is defined as the property of a substance which becomes evident during chemical change in which a change in chemical composition takes place. A new substance is formed in these reactions.

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Phosphorus will ignite when exposed to air: It undergoes oxidation which is a chemical change

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