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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Answer:
its a sure for not but i think its a 192583
Answer:
<h3>1/16</h3>
Explanation:
According to the coulombs law, the force existing vetween the ions is expressed as;
F = kQq/r² .... 1
Q and q are the ions
r is the distance between the ions
If the distance between the ion is quadrupled, then;
F2 = kQq/(4r)²
F2 = kQq/16r² ... 2
Divide equation 2 by 1;
F2/F = kQq/16r² ÷ kQq/r²
F2/F = kQq/16r² × r²/kQq
F2/F = 1/16
F2 = 1/16 F
Therefore the coulombic force between two ions is reduced to<u> 1/16 </u>of its original strength when the distance between them is quadrupled.
