(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
2 answers:
Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)
Answer:
d= 0.46 m
Explanation:
The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.
For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:
V =
We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.
If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.
So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:
V = +
= 0
⇒:
Replacing by the values of q1, q2, and k, and solving for x, we get:
⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.
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Answer:
a) attractiva, b) dF = , c) F =
, d) F = -1.09 N
Explanation:
a) q1 is negative and the charge of the bar is positive therefore the force is attractive
b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x
dF =
where k is a constant, Q₁ the charge at the origin, x the distance
c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L
∫ dF =
as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density
λ = dQ₂ / dx
DQ₂ = λ dx
we substitute
F =
F = k Q1 λ ()
we evaluate the integral
F = k Q₁ λ
F = k Q₁ λ
we change the linear density by its value
λ = Q2 / L
F =
d) we calculate the magnitude of F
F =9 10⁹ (-4.2 10⁻⁶)
F = -1.09 N
the sign indicates that the force is attractive
Answer:
x = 129.9 m
y = 30.9 m
Explanation:
When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.
Given data:
= 50 m/s
Angle = 30°
Time = t = 3 s
horizontal component of velocity = =
cos30°
= 50cos30°
= 43.3 m/s
Vertical component of velocity = =
Sin30°
= 50Sin30°
= 25 m/s
This is a projectile motion, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.
But the vertical component of velocity varies with time and there is an acceleration along vertical direction which is equal to gravitational acceleration g.
Horizontal distance = x = t
x = 43.3*3
x = 129.9 m
Vertical Distance = y = t -0.5gt²
y = 25*3 - 0.5*9.8*3²
y = 75 - 44.1
y = 30.9 m
The ball will take 0.45 seconds to reach the ground
<h3>How will we solve this question?</h3>We will use Newton’s laws of motion, so
H= taking height= 1m
The 3 laws of motion given by Newton are as follows:
1) Unless an unbalanced force acts upon it, an item at rest remains at rest, and an object in motion continues to move in a straight line at a constant pace.
2) An object's acceleration is influenced by its mass and the force being applied.
3) Whenever one thing applies force to another, the second object applies the opposing, equal force to the first.
To know more about Newton's laws of motion visit:
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