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mars1129 [50]
2 years ago
11

A positively-charged particle is released near the positive plate of a parallel plate capacitor. a. Describe its path after it i

s released and explain how you know. b. If work is done on the particle after its release, is the work positive or negative
Physics
1 answer:
il63 [147K]2 years ago
8 0

Answer:

a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then<em> it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq</em>, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to

This can be calculated by Gauss' Law.

A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.

b. The particle moves from the higher potential to the lower potential. <em>The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.</em>

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A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t
Pie

Answer:

I = \frac{mvb}{6}

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

       L = Iω ....                        (1)

moment of inertia of rod rotating about its center of length b

 

      I = \frac{ mb^2}{12}  ........               .(2)  

using         v = ωr  

where w is angular velocity

and r is radius of  rod which is equal to b

        so we get  2v =  ωb  

                            ω  = 2v/b  .................            (3)    

here velocity is two time because two opposite ends  are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

                 L   =   \frac{mb^2}{12}×\frac{2v}{b}

              so final answer will be      L  =   \frac{mvb}{6}

7 0
2 years ago
A nonrelativistic electron is accelerated from rest through a potential difference. After acceleration the electron has a de Bro
aleksley [76]

Answer:

Potential difference though which the electron was accelerated is 2.67\times 10^{-6}\ uV\  .

Explanation:

Given :

De Broglie wavelength , \lambda=750\ nm.

Plank's constant , h=6.626\times 10^{-34}\ J.s \ .

Charge of electron , e=-1.6\times 10^{-19}\ C.

Mass of electron , m=9.11\times 10^{-31}\ kg.m=9.11\times 10^{-31}\ kg.

We know , according to de broglie equation :

\lambda=\dfrac{h}{mv}\\\\ v=\dfrac{h}{m\lambda}\\\\v=\dfrac{6.626\times 10^{-34}\ J.s \ }{9.11\times 10^{-31}\ kg\times 750\times 10^{-9}\ m }= 969.78\ m/s .

Now , we know potential energy applied on electron will be equal to its kinetic energy .

Therefore ,

qV=\dfrac{mv^2}{2}\\\\ V=\dfrac{mv^2}{2q}

Putting all values in above equation we get ,

V=2.67\times 10^{-6}\ uV .

Hence , this is the required solution.

5 0
2 years ago
A ball on a string travels once around a circle with a circumference of 2.0 m. The tension in the string is 5.0 N. how much work
SpyIntel [72]

Answer:0

Explanation:

Given

circumference of circle is 2 m

Tension in the string T=5 N

2\pi r=2

r=\frac{2}{2\pi }=\frac{1}{\pi }=0.318 m

In this case Force applied i.e. Tension is Perpendicular to the Displacement therefore angle between Tension and displacement is 90^{\circ}

W=\int\vec{F}\cdot \vec{r}

W=\int Fdr\cos 90

W=0

4 0
1 year ago
A metal sphere of radius 10 cm carries a charge of +2.0 μC uniformly distributed over its surface. What is the magnitude of the
frozen [14]

Answer:

8.0\cdot 10^5 N/C

Explanation:

Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.

Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:

E=k\frac{q}{(R+r)^2}

where

k=8.99\cdot 10^9 N m^2C^{-2} is the Coulomb's constant

q=2.0 \mu C=2.0 \cdot 10^{-6}C is the charge on the sphere

R=10 cm = 0.10 m is the radius of the sphere

r=5.0 cm = 0.05 m is the distance from the surface of the sphere

Substituting, we find

E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C

3 0
2 years ago
Which statement about images is correct? a) A virtual image cannot be formed on a screen. b) A virtual image cannot be viewed by
ankoles [38]

Answer:

A). A virtual image cannot be formed on a screen.

Explanation:

A virtual image can not be formed on a screen.

For image:

1.A virtual image can be viewed by the unaided eye.

2. A real image must be erect or maybe inverted.

3.Mirrors can produce virtual as well as real image ,it depends on which type of mirror is.

4.A virtual image can be photographed.

So the option A is correct.

5 0
2 years ago
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