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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.
The object's <u>initial velocity</u> is equal to
Why?
From the statement we know the height of the tree and the time it takes to reach the ground, so, if we need to calculate its initial velocity, we can use the following formula:
Where,
y, is the final height (0 meters in this case)
yo, is the initial height (111 meters in this case)
t, is the time elapsed (3.8 seconds in this case)
vo, is the initial speed.
g, is the acceleration due to gravity (-9.81 m/s2)
Now, let's set the origin at the top of the tree, so, rewriting the formula, we have:
So, isolating the initial velocity, we have:
Finally, substituting and calculating, we have:
Hence, we have that the <u>initial velocity</u> of the object is
Have a nice day!