The electric field strength of a point charge is inversely proportional to the square of the distance from the charge ... a lot like gravity.
If the magnitude of the field is (2E) at the distance 'd', then at the distance '2d', it'll be (2E)/(2²). That's (2E)/4 = 0.5E .
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Answer:
The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts
Explanation:
Electric potential is given as;
V = E*r
where;
E is the electric field strength, = kq/r²
V = ( kq/r²)*r
V = kq/r
k is coulomb's constant = 8.99 X 10⁹ Nm²/C²
q is the charge of the particles = 1.6 X 10⁻¹⁹ C
r is the distance between the particles = 859 nm
At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm
V = (8.99 X 10⁹ * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)
V = 3.349 X 10⁻³ Volts
Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts
A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,


B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

where
are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that
, which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual