Chapter 20, Problem 61A

(a)

To determine

The position of a third sphere $C$ so that there will be zero net force on it.

Answer to Problem 61A

  $d_{AC}\text{ = +2}\text{.0 m}$ on the x -axis

Explanation of Solution

Given:

The charge on sphere $A$ is $q_A\text{ = 64 }\mu \text{C = 64}\times {\text{10}}^{-6}\text{ C}$

The position of the charge $A$ is origin

The charge on sphere $B$ is $q_B\text{ = -16 }\mu \text{C = -16}\times {\text{10}}^{-6}\text{ C}$

The position of the charge $B$ is $\text{+1}\text{.0 m}$ on the x-axis

Formula used:

Coulomb’s law: If $A$ and $B$ are the two objects having charges $q_A$ and $q_B$ respectively which are separated by a distance $d$ , then Coulomb’s law can be written as,

  $F\text{ = K}\frac{q_A{q}_B}{d^2}$

Where $\text{ K}$ is a constant, $\text{ K}=9\times {10}^9\text{ N}{\text{.m}}^2/C^2$

Calculation:

For the zero-net force on sphere $C$ , the attractive and repulsive forces must cancel each other. That means,

  $F_{AC}=\text{K}\frac{q_A{q}_C}{d_{AC}^2}=\text{K}\frac{q_B{q}_C}{d_{BC}^2}=F_{BC}$

  $\Rightarrow \frac{q_A}{d_{AC}^2}=\frac{q_B}{d_{BC}^2}$

  $\Rightarrow q_A{d}_{BC}^2=q_B{d}_{AC}^2$

  $\Rightarrow d_{AC}^2=\frac{q_A{d}_{BC}^2}{q_B}$

  $\Rightarrow d_{AC}=\sqrt{\frac{q_A{d}_{BC}^2}{q_B}}$ $\left(1\right)$

Where, $q_C$ is the charge on sphere $C$

$d_{AC}$ is the distance between the sphere $A$ and $C$

$d_{BC}$ is the distance between the sphere $B$ and $C$

The position of third sphere $C$ is,

  $\begin{array}{c}{d}_{AC}=\sqrt{\frac{\left(\text{64}\times {\text{10}}^{-6}\text{ C}\right)d_{BC}^2}{\left(\text{16}\times {\text{10}}^{-6}\text{ C}\right)}}\\ =+2d_{BC}\\ =+2\times 1\\ =+2.0\text{m}\end{array}$

Conclusion:

The third sphere $C$ should be placed at $\text{+2}\text{.0 m}$ on the x-axis so that there is no net force on it.

(b)

To determine

The position of a third sphere $C$ for the given charge value.

Answer to Problem 61A

  $d_{AC}\text{ = +2}\text{.0 m}$ on the x -axis

Explanation of Solution

Given:

The charge on sphere $A$ is $q_A\text{ = 64 }\mu \text{C = 64}\times {\text{10}}^{-6}\text{ C}$

The position of the charge $A$ is origin

The charge on sphere $B$ is $q_B\text{ = -16 }\mu \text{C = -16}\times {\text{10}}^{-6}\text{ C}$

The position of the charge $B$ is $\text{+1}\text{.0 m}$ on the x -axis

The charge on sphere C is $q_C\text{ = 16 }\mu \text{C = 16}\times {\text{10}}^{-6}\text{ C}$

Formula used:

Coulomb’s law: If $A$ and $B$ are the two objects having charges $q_A$ and $q_B$ respectively which are separated by a distance $d$ , then Coulomb’s law can be written as,

  $F\text{ = K}\frac{q_A{q}_B}{d^2}$

Where $\text{ K}$ is a constant, $\text{ K}=9\times {10}^9\text{ N}{\text{.m}}^2/C^2$

Calculation:

For the zero-net force on sphere $C$ , the attractive and repulsive forces must cancel each other. That means,

  $F_{AC}=\text{K}\frac{q_A{q}_C}{d_{AC}^2}=\text{K}\frac{q_B{q}_C}{d_{BC}^2}=F_{BC}$

  $\Rightarrow \frac{q_A}{d_{AC}^2}=\frac{q_B}{d_{BC}^2}$

  $\Rightarrow q_A{d}_{BC}^2=q_B{d}_{AC}^2$

  $\Rightarrow d_{AC}^2=\frac{q_A{d}_{BC}^2}{q_B}$

  $\Rightarrow d_{AC}=\sqrt{\frac{q_A{d}_{BC}^2}{q_B}}$ $\left(1\right)$

Where, $q_C$ is the charge on sphere $C$

$d_{AC}$ is the distance between the sphere $A$ and $C$

$d_{BC}$ is the distance between the sphere $B$ and $C$

It is observed that the charge $q_C$ gets cancel from the equation. So, the position of sphere $C$ is independent of charge and sign of the sphere $C$ .

So, the position of third sphere $C$ will be

  $d_{AC}=+2.0\text{m}$