Chapter 17, Problem 22P

(a)

To determine

The point along x-axis at which electric field is zero.

Answer to Problem 22P

The answer is $22cmleftof{q}_2$ .

Explanation of Solution

Given data:

Charge $q_1=3\mu C$

Charge $q_2=-2\mu C$

Formula used:

It is known that: $E=k\frac{\left|q_2\right|}{x^2}-k\frac{q_1}{{\left(d+x\right)}^2}=0$

Where,

  $k$ is the constant factor.

  ${\text{q}}_1\&{\text{q}}_2$ are the charges.

  $\text{d}$ is the distance between two charges.

Calculation:

Due to an inverse square nature of the electric field it is zero at any location that is closer to weaker charge. In following diagram the point ‘x’ is taken as where charge are weaker. Taking right as the positive direction

Rewrite the formula $E=k\frac{\left|q_2\right|}{x^2}-k\frac{q_1}{{\left(d+x\right)}^2}=0$ ,

  $x=\frac{\sqrt{\left|q_2\right|}}{\sqrt{q_1}-\sqrt{\left|q_2\right|}}d$

Substitute the data,

  $\therefore x=\frac{\sqrt{2.0\times {10}^{-6}C}}{\sqrt{3.0\times {10}^{-6}C}-\sqrt{3.0\times {10}^{-6}C}}\left(5.0cm\right)$

  $\therefore x=\frac{\sqrt{2.0\times {10}^{-6}C}}{\sqrt{3.0\times {10}^{-6}C}-\sqrt{3.0\times {10}^{-6}C}}\left(5.0cm\right)$

  $\therefore x=22cmleftof{q}_2$

Conclusion: $22cmleftof{q}_2$ at this point electric field is zero.

(b)

To determine

The point along x-axis at which the potential is zero.

Answer to Problem 22P

The answers are $2.0cm$ and $10.0cm$ .

Explanation of Solution

Given data:

Charge $q_1=3\mu C$

Charge $q_2=-2\mu C$

Voltage, $V=0$

Resistance, $r=\infty$

Formula used:

It is known that: $x=\frac{q_{2}d}{\left(q_2-q_1\right)}$

Where,

  ${\text{q}}_1\&{\text{q}}_2$ are the charges.

  $\text{d}$ is the distance between two charges.

  $x$ is the required location.

Calculation:

Since potential is zero, hence the positive charge is must be closer to charge. Let $x_1$ be the position between charges and position of $x_2$ be left to the negative charge as shown in diagram.

Apply the formula and substitute the data $x=\frac{q_{2}d}{\left(q_2-q_1\right)}$

  $\Rightarrow x_1=\frac{\left(-2.0\times {10}^{-6}C\right)\left(5.0cm\right)}{\left(-5.0\times {10}^{-6}C\right)}$

  $\Rightarrow x_1=2.0cm$

  $\Rightarrow x_2=\frac{\left(-2.0\times {10}^{-6}C\right)\left(5.0cm\right)}{\left(-1.0\times {10}^{-6}C\right)}$

  $\Rightarrow x_2=10.0cm$

Conclusion: At both locations $2.0cm$ and $10.0cm$ the potential is zero.