Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.
<h3>What is electric force?</h3>
Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.
The electric force in the second case will be the same as in the first case. Therefore the force on the particle will be the same.
Hence the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.
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Answer:
Explanation:
The relationship between the refractive index and the critical angle is given as follows:
where,
η = refractive index = 1.67
θc = critical angle =?
Therefore,
Answer:
B. It is directly proportional to the source charge.
Explanation:
Gauss's law states that the total (net) flux of an electric field at points on a closed surface is directly proportional to the electric charge enclosed by that surface.
This ultimately implies that, Gauss's law relates the electric field at points on a closed surface to the net charge enclosed by that surface.
This electromagnetism law was formulated in 1835 by famous scientists known as Carl Friedrich Gauss.
Mathematically, Gauss's law is given by this formula;
ϕ = (Q/ϵ0)
Where;
ϕ is the electric flux.
Q represents the total charge in an enclosed surface.
ε0 is the electric constant.
Hence, the statement which is true of the electric field at a distance from the source charge is that it is directly proportional to the source charge.
Answer:
The energy of the capacitors connected in parallel is 0.27 J
Given:
C = 
C' = 
Potential difference, V = 300 V
Solution:
Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:
The energy of the capacitor, E is given by;
Answer:
The object will move to Xfinal = 7.5m
Explanation:
By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:
Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s
With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:
Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =
= 3.9m + 3.6m = 7.5m
