The charge on the particle is 5.6 × 10⁻¹¹ C.
<h3>Calculation:</h3>
The magnitude of an electric field produced by a charge is given by:
E = q/ 4πε₀r²
where,
E = electric field
q = charge
r = distance
1/4πε₀ = 8.99 × 10⁹ Nm²/C²
Given,
E = 2.0 N/C
r = 50 cm = 0.5 m
To find,
q =?
Put the values in the above equation:
E = q/ 4πε₀r²
q = E (4πε₀r²)
q = 2.0 × (0.50²)/ 8.99 × 10⁹
q = 5.6 × 10⁻¹¹ C
Therefore, the particle has a charge of 5.6 × 10⁻¹¹ C.
<h3>What is an electric field?</h3>
The physical field that surrounds each electric charge and acts to either attract or repel all other charges in the field is known as an electric field. Electric charges or magnetic fields with different amplitudes are the sources of electric fields.
I understand the question you are looking for is this:
A charged particle produces an electric field with a magnitude of 2.0 N/C at a point that is 50 cm away from the particle. What is the magnitude of the particle's charge?
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A star with greater mass will die out faster than the Sun.
<h3>What factors star is dependent on?</h3>
A star's future relies upon its mass. For the most part, the more huge the star, the quicker it consumes its fuel supply, and the more limited its life. The most huge stars can wear out and detonate in a cosmic explosion after two or three million years of combination.
Our Sun is a typical estimated star: there are more modest stars and bigger stars, even up to multiple times bigger. Numerous other planetary groups have different suns, while our own simply has one. The Sun is made for the most part out of hydrogen and helium gas.
In this manner, one correlation in the occasions in the existence of the Sun with those of a star that beginnings with a mass multiple times more prominent than the Sun's is a star that has a more noteworthy mass will vanish quicker.
Learn more about Star.
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Answer:
3 m/s
Explanation:
<u>The velocity of a position-time graph is the slope of the line.</u> Slope is rise over run, or rise divided by run. The rise (how many units the line goes up) is 3 units and the run is 1 unit. 3/1 is 3, so the velocity is 3 m/s.
Answer:
B. w=12.68rad/s
C. α=3.52rad/s^2
Explanation:
B)
We can solve this problem by taking into account that (as in the uniformly accelerated motion)
( 1 )
where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.
In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have
to calculate the angular speed w we can use
Thus, wf=12.68rad/s
C) We can use our result in B)
I hope this is useful for you
regards
Answer:
A. 
B. 
C. 
Explanation:
The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is
is the capacitance, 
is the common plate area, 
is the plate separation and 
is the permittivity of the material between the plates.
For air or free space, is 
called the permittivity of free space. In general, 
where 
is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, 
.
The energy stored in a capacitor is the average of the product of its charge and voltage.
Its charge, 
, is related to its capacitance by 
(this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for 
,
A. Substituting for in ,
B. When the distance is 
,
C. When the distance is restored but with a dielectric material of dielectric constant, 
, inserted, we have
