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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0.120L + 2.345L = 2.465L = 4 significant figures in the answer
A lab cart is loaded with different masses and moved at various constant velocities? the anser should be
1.0m/s → 4kg
Answer:4.34 miles
Explanation:
first Elevation =
After 1 minute Elevation changes to 
Ditsance travelled in 1 minute =
=10 mile
Now
tan59=
H=xtan59
tan19=
H=
Equating H
we get
1.319x=10tan19
x=2.61 miles
H=
=4.34 miles
Answer:
The induced current can be increased in the coil in the following ways: By increasing the strength of the magnet. By increasing the speed of the magnet through the coil.
Explanation:
