In general, we know that the force of the electric field exerted on a point charge q at distance r from charge Q is:
F = k Qq/r²
If you have more than one charge, then the total force is the vectorial sum of the forces created by each charge.
Moreover, two charges with the same sign repulse each other, while two charges with opposite sign attract each other. This is fundamental to understand the direction of the force.
We will define "to the right" the direction towards increasing positive values of the x-axis (and we will assign a positive value), and "to the left" the direction towards decreasing negative values of the x-axis of x <span>(and we will assign a negative value).
Before considering each position, it is better to transform our data into the correct units of measurements:
q</span>₁ = 6.0×10⁻⁶C
q₂ = 1.5×10⁻⁶C
q₃<span> = 2.0×10⁻⁶C
d</span>₂ = 3×10⁻²m
d₃= 5×10⁻²m
A) In position 1, we have a positive charge (q₁) on which is exerted a repulsive force by another positive charge (q₂) - which will be to the left because the charge q₁ will be pushed away- and an attractive force by a negative charge - which will be to the right:
F₂₁ = 9×10⁹ · 1.5×10⁻⁶ · 6.0×10⁻⁶ / (3×10⁻²)² = 90N
F₃₁ = 9×10⁹ · 2.0×10⁻⁶ · 6.0×10⁻⁶ / (5×10⁻²)<span>² = 43.2N
The total force exerted on q</span>₁ will be:
F₁ = -90 + 43.2 = - 46.8N (negative, then to the left)
B)In position 2, we have a positive charge (q₂) on which is exerted a repulsive force by another positive charge (q₁) - which will be to the right because the charge q₂ will be pushed away- and an attractive force by a negative charge <span>(q₃)</span> - which will be to the right. We expect F₁₂ to be equal in magnitude but opposite to F₂₁ found in point A):
F₁₂ = 9×10⁹ · 6.0×10⁻⁶ · 1.5×10⁻⁶/ (3×10⁻²)² = 90N
F₃₂ = 9×10⁹ · 2.0×10⁻⁶ · 1.5×10⁻⁶ / (2×10⁻²)<span>² = 67.5N
The total force exerted on q</span>₂ will be:
F₂ = 90 + 67.5 = 157.5N (positive, then to the right)
C) In position 3, we have a negative charge (q₃) on which is exerted an attractive force by a positive charge (q₁) - which will be to the left - and an attractive force by another positive charge (q₂) - which will be to the left. We expect F₁₃ = -F₃₁ and F₂₃ = -F₃₂:
F₁₃ = 9×10⁹ · 6.0×10⁻⁶ · 2.0×10⁻⁶/ (5×10⁻²)² = 43.2N
F₂₃ = 9×10⁹ · 1.5×10⁻⁶ · 2.0×10⁻⁶ / (2×10⁻²)<span>² = 67.5N
The total force exerted on q</span>₂ will be:
F₃ = -43.2 - 67.5 = -110.7N (negative, then to the left)
Answer:
distance between object and image = 18.9 cm
Explanation:
given data
radius of curvature = 18 cm
focal length = 1/2 radius of curvature
magnification = 40%
to find out
distance between object and image
solution
we know lens formula that is
1/f = 1/v + 1/u ....................1
here f = 18 /2 and v and u is object and image distance
and we know m = 40% = 0.40
so 0.40 = -v / u
so here v = - 0.40 u
so from equation 1
1/f = 1/v + 1/u
2/18 = - 1/0.40u + 1/u
u = -13.5 cm ..................2
and
v = -0.40 (- 13.5)
v = 5.4 cm ......................3
so from equation 2 and 3
distance between object and image = 5.4 + 13.5
distance between object and image = 18.9 cm
Hello!
Yes, someone with an IQ score of 120 can be gifted.
An IQ score of 120 means that the individual is moderately gifted. Although this IQ doesn't qualify for gifted education programs, research shows that people with IQs of about 145-180 don't perform significantly better than people with 120s IQ in terms of success.
As an example, professor Dean Simonton from the University of California found that an optimum leader should have an IQ in the 120-125 range, so having an IQ of 120 would mean that the individual is gifted for leadership.
Have a nice day!
They can be compounds too.
Examples: salt, sugar, and water are all compounds, not elements or mixtures.
675 nM is 675 x 10⁻⁹ meter ... 0.000 000 675 meter .
That's 0.000 067 5 of one centimeter
0.000 675 of one millimeter