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?
Learn more about electric field here:
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<span>when the gravitational forces and air resistance equalize on an object that is falling toward the earth and the object stops accelerating its called TERMINAL VELOCITY</span>
Answer:
It gets refracted.
Explanation:
When light beam travels through different mediums they refract i.e. they change their direction. Here the angle of incidence is less than 90°. After entering the glass slab the light beam will move towards the normal (a line drawn perpendicular to the interface of the two mediums). Thus the angle of refraction will be even lesser than angle of incidence.
