The work done by the electric field in moving a charge is the negative of the potential energy difference between the two locations, which is the product between the magnitude of the charge q and the potential difference
:
The proton charge is
, and the two locations have potential of
and
, therefore the work is
Answer:
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
Explanation:
The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.
The first minimum occurs for m = 1, so the diffraction equation of a slit remains
a sin θ = λ
in general, the diffraction patterns occur at very small angles, so
sin θ = θ
θ = λ / a
in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.
θ = 1.22 λ /a
In this exercise we are told that the opening changes
a’ = 2 a
we substitute
θ ‘= 1.22 λ / 2a
θ' = (1.22 λ / a) 1/2
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
Answer:
C
Sign-Negative
Explanation:
We are given that
Electric field =
(Radially downward)
Acceleration=
(Upward)
Mass of charge=3 g=
kg
1kg=1000g
We have to find the magnitude and sign of charge would have to be placed on a penny .
By newton's second law
Substitute the values then we get
Substitute the values then we get
C
Sign of charge =Negative
Because electric force acting in opposite direction of electric field therefore,charge on penny will be negative.
I believe the answer should be the last option. upon interaction, both objects should have the same charge after the electrons are transferred.
