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1 answer:
The intensity of the electric field is 30,000 N/C
Explanation:
The strength of the electric field produced by a single-point charge is given by the equation
where:
is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the magnitude of the charge
r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity
Substituting, we find:
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Answer:
angular range is ( 0.681 rad , 0.35 rad )
Explanation:
given data
wavelength λ = 380 nm = 380 × m
wavelength λ = 700 nm = 700 × m
to find out
angular range of the first-order
solution
we will apply here slit experiment equation that is
d sinθ = m λ ...........1
here m is 1 for single slit and d is =
so put here value in equation 1 for 380 nm
we get
d sinθ = m λ
sinθ = 1 × 380 ×
θ = 0.35 rad
and for 700 nm
we get
d sinθ = m λ
sinθ = 1 × 700 ×
θ = 0.681 rad
so angular range is ( 0.681 rad , 0.35 rad )