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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ans D. Substance B has a greater latent heat of vaporization than substance A.
Answer:
0.25 m
Explanation:
We can solve the problem by using the lens equation:
where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem, we have
f = +20 cm=+0.20 m (the focal length is positive for a converging lens)
q = +1.0 m (the image distance is positive for a real image)
Solving the equation for p, we find
Answer:
1.6 x 10^-5 T
Explanation:
i = 4 A
r = 0.05 m
The magnetic field due to long wire at a distance r is given by
B = 10^-7 x 2 x 4 / 0.05
B = 1.6 x 10^-5 T
During the collision between two balls on the pool table there is no external force along the line of collision between them
Since there is no external force on it so here we can say
here we have
so we can say
since there is no external force so we can say during the collision the momentum of two balls will remain conserved
