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:
brainly.com/question/14857134
#SPJ4
Answer:
46
Explanation:
HOPE IT'S HELPFUL FOR U MATE AND PLZ MARK AS BRAINLIEST I REALLY NEED IT
Answer:
yes
Explanation:
please Mark my answer in brainlist and follow me
Answer:
Fn: magnitude of the net force.
Fn=30.11N , oriented 75.3 ° clockwise from the -x axis
Explanation:
Components on the x-y axes of the 17 N force(F₁)
F₁x=17*cos48°= 11.38N
F₁y=17*sin48° = 12.63 N
Components on the x-y axes of the the second force(F₂)
F₂x= −19.0 N
F₂y= 16.5 N
Components on the x-y axes of the net force (Fn)
Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N
Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N
Magnitude of the net force.
Direction of the net force (β)
β=75.3°
Magnitude and direction of the net force
Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis
In the attached graph we can observe the magnitude and direction of the net force
