What is the range of a ball thrown horizontally at 12 m/s if its time of flight is 3.0 s?
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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.
By using Coulomb's law, we want to find the value of q₁ given that q₂ experiences no net electric force. We will find that q₁ = 8nC
<h3>Working with Coulomb's law.</h3>Coulomb's law says that for two charges q₁ and q₂ separated by a distance r, the force that each one experiences is:
Where k is a constant
Here we can see that q₂ interacts with two charges, then the total force on q₂ will be:
And we know that it must be equal to zero, so we can write it as:
The parenthesis must be equal to zero, so we can write:
And now we can solve this for q₁ to get:
If you want to learn more about Coulomb's law, you can read: