Answer: The correct answer is- Field.
Explanation -
The modified space around a charged particle is known as electric field.
The strength of electric field at a point is calculated as the force experienced by a unit test charge present at that point.
Thus, a region of space in which a test charge experiences a force is referred to as Field.
Answer:
(a) Charge density σ=6.6375×10²nC/m²
(b) Total charge Q=1.47×10²nC
Explanation:
Given Data
A=47.0 cm =0.47 m
Electric field E=75.0 kN/C
To find
(a) Charge density σ
(b)Total Charge Q
Solution
For (a) charge density σ
From Gauss Law we know that
Φ=Q/ε₀.......eq(i)
Where
Φ is electric flux
Q is charge
ε₀ is permittivity of space
And from the definition of flux
Φ = EA
The flux is electric field passing perpendicularly through the surface
Put the this Φ in equation(i)
EA
=Q/ε₀
where Q(charge)=σA
EA=(σA)/ε₀
E=σ/ε₀
σ=ε₀E

σ=6.6375×10²nC/m²
For (b) total charge Q
Q=σA

Answer:
Kinetic energy = 127.89 Joules
Explanation:
Kinetic energy is calculated using the following rule:
KE = (1/2)*v* v^2
Where:
m is the mass = 145 g = 0.145 kg
v is the velocity = 42 m/sec
Substitute in the above equation to get the kinetic energy as follows:
KE = (0.5)(0.145)(42)^2
Kinetic energy = 127.89 Joules
Hope this helps :)