Question

(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 1

Answer:

The question seems to be incomplete, but ill try to answer it the best that I can do it.

We have two charges in a horizontal line, where the charge of the charges is q1 and q2, and the distance in between the charges is d. So we can think that one of the charges is in x = 0 and the other charge is in x = d, and we can solve it only on x.

We want to find a point in between the charges where the electric potential is zero.

The electric potential for a charge is:

V(x) = k*q/Ix - x0I

where x0 is the position of the charge, and k is a constant.

Then the potentials that we have aer

V1(x) = k*q1/Ix-0I = k*q1/IxI

V2(x) =  k*q1/IxI

now we want to find for wich value of x V1(x) + V2(x) = 0

k*q1/IxI +  k*q2/Ix - dI = 0

k(q1/IxI +  q2/Ix - dI) = 0  

So here we can see that the signs of q1 and q2 must be different:

q1/IxI = -q2/Ix-dI

q1*Ix-dI = -q2*IxI

With this relation, you can find x.

I can not keep solving it without the values of q1, q2, and d, but with this equation, you can find the value of x.

Answer 2

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 = $\frac{k*q}{r}$

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 = $\frac{k*q1}{x}$ +$\frac{k*q2}{(1.63m-x)}$ = 0

⇒$k*q1* (1.63m - x) = -k*q2*x$:

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.