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AysviL [449]
3 years ago
12

Consider a disk of radius 4.9 cm, having a uniformly distributed charge of +6.1 μC. The value of the Coulomb constant is 8.98755

x 10 N m2/C2 Compute the magnitude of the electric field at a point on the axis and 3.1 mm from the center Answer in units of N/C.

Physics
1 answer:
gayaneshka [121]3 years ago
5 0

Answer:

the electric field is 42.8×10^6 N/C

Explanation:

let k be the coulomb constant, Q be the charge distribution , R be the radius of the disk and x be the distance from the center.

k = 8.98755×10^9 N×m^2/c^2

Q = 6.1×10^-6 C

R = 4.9×10^-2 m

x = 3.1×10^-3 m

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Newton's third law states "for every action, there is an equal and opposite reaction."

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Answer:

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Explanation:

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4 0
2 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
Imagine a system where a block rests on an inclined plane. The block is then given an initial push so that it starts sliding dow
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Answer:

statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.

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The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.

The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction

here, the downward direction signifies the downward motion parallel to the inclined plane.

Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.

Hence, for the block to stop sliding the the above statement should be true.

6 0
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Solid and liquids are much more denser than gas because their molecules are close to each other and with that the molecules of them can't move that freely unlike the gas molecules. Also, because of being near to each other the molecules of solid and liquids became heavy making them dense.
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