Question

A charge q is at the point x = 5.0 m , y = 0. Write expressions for the unit vectors you would use in Coulomb's law if you were finding the force that q exerts on other charges located at A) x1=5.0m, y1=1.0m B) the origin C) x2=4.5m, y2=3.5m.

Answer 1

Answer:

A)    F = kq2 ( j ^) N

, B)    F = kq² / 25 (i ^) N, C)     F = kq²  (4 i ^ + 0.0816 j ^)  N

Explanation:

Coulomb's Law is

           F = k q₁q₂ / r²

Where k is the Coulomb constant that is worth 8.99 109 N m2 / C2, what are the charges and r the distance between them

 

This force can be attractive or repulsive, depending on the sign of the charges, if they are of the same sign the force is repulsive.

A) the charges are at points (5.0, 0) and (5.0, 1.0) the value of the charges is q and for being of the same sign the force is repulsive

Let's look for the distance

              r² = (x2-x1) 2 + (y2-y1) 2

              r² = (5-5) 2 + (1-0) 2

             r² = 1 m

We see that the distance is in the direction of the y axis, so the force eta in this direction

Strength is

               F = K q² / 1 j ^

              F = kq² (0 i ^ + j ^) N

B) the ₈second load at the origin (0, 0)

               r² = (5-0)² + (0-0)²

               r² = 25

                 

              F = k q² / 25 i ^

             F = kq² / 25 (i ^ + 0 j ^) N

C) the second charge at point (4.5, 3.5)

Since we have values ​​on both axes, the best method is to look for the force components on each axis, since the total force is in the direction of the distance vector,

X axis

        Fₓ = k q² / (4.5-5)²

        Fₓ = k q² / 0.25 = k q² 4

Y Axis

        = k q² / (3.5-0)²

        = k q² / 12.25 = k q²    0.0816

       F = kq²  (4 i ^ + 0.0816 j ^)  N

Answer 2

Answer:

A) $\^r = \^y$

B) $\^r = -\^x$

C) $\^r = -0.14\^x + 0.99\^y$

Explanation:

Coulomb's Law is

$\vec{F}_{ab} = K\frac{q_1q_2}{r^2}\^r$

where r is the distance between the two point charges.

The question clearly asks the unit vector expressions of the force that q exerts on the other charge. So, we do not need to find the force, but only the distance vector, r, between charges, and then we can derive the unit vector pointing only the direction of the force.

A) $\vec{r} = \vec{r}_b - \vec{r}_a = (5\^x + \^y) - (5\^x + 0) = \^y\\\^r = \frac{\vec{r}}{|\vec{r}|} = \frac{\^y}{1} = \^y$

B) $\vec{r} = \vec{r}_b - \vec{r}_a = (0 + 0) - (5\^x + 0) = -5\^x\\\^r = \frac{\vec{r}}{|\vec{r}|} = \frac{-5\^x}{5} = -\^x$

C) $\vec{r} = \vec{r}_b - \vec{r}_a = (4.5\^x + 3.5\^y) - (5\^x + 0) = -0.5\^x + 3.5\^y\\\^r = \frac{\vec{r}}{|\vec{r}|} = \frac{-0.5\^x + 3.5\^y}{\sqrt{(0.5)^2+(3.5)^2}} = \frac{-0.5\^x + 3.5\^y}{3.53} = -0.14\^x + 0.99\^y$