Chapter 17, Problem 63GP

(a)

To determine

The points along the line where the electric field is zero.

Answer to Problem 63GP

  $\text{11}\text{.1}\text{cm}$ from the negative charge.

Explanation of Solution

Given:

The point charges are $\text{3}\text{.4μC}\text{=}\text{3}\text{.4}\text{×}{\text{10}}^{\text{-6}}\text{C}$ and $\text{-2}\text{.6μC}\text{=}\text{2}\text{.6}{\text{×10}}^{\text{-6}}\text{C}$ respectively and are placed $\text{1}\text{.6}\text{cm}$ apart.

Formula used:

The magnitude of electric field $\text{E}$ due to a point charge $\text{q}$ at a distance $\text{r}$ from the charge is given as

  $\text{E=k}\frac{\text{q}}{{\text{r}}^{\text{2}}}$

Where, $\text{k}$ is the Coulomb’s constant.

Calculation:

The electric field is zero at points where the electric field of the two charges cancel each other.

  ${\text{E}}_{\text{1}}{\text{=E}}_{\text{2}}$

  $\text{k}\frac{{\text{q}}_{\text{1}}}{{\left(\text{d+r}\right)}^{\text{2}}}\text{=k}\frac{{\text{q}}_{\text{2}}}{{\left(\text{d+r}\right)}^{\text{2}}}$

According to the question:

  $\text{E =}\text{k}\frac{\left|{\text{q}}_{\text{2}}\right|}{{\text{r}}^{\text{2}}}\text{-k}\frac{{\text{q}}_{\text{1}}}{{\left(\text{d+r}\right)}^{\text{2}}}\text{=0}\to \left|{\text{q}}_{\text{2}}\right|{\left(\text{d+r}\right)}^{\text{2}}{\text{=q}}_{\text{1}}{\text{r}}^{\text{2}}$

  $\text{r}\text{=}\frac{\sqrt{\left|{\text{q}}_{\text{2}}\right|}}{\sqrt{{\text{q}}_{\text{1}}}\text{-}\sqrt{\left|{\text{q}}_{\text{2}}\right|}}\text{d}$

  $\text{=}\frac{\sqrt{\text{2}{\text{.6×10}}^{\text{-6}}\text{C}}}{\sqrt{\text{3}{\text{.4×10}}^{\text{-6}}\text{C}}\text{-}\sqrt{\text{2}{\text{.6×10}}^{\text{-6}}\text{C}}}\left(\text{1}\text{.6}\text{cm}\right)$

  $\text{r }\text{=}\text{11}\text{.1}\text{cm}$ left of ${\text{q}}_{\text{2}}$

(b)

To determine

The points along the line where the electric potential is zero.

Answer to Problem 63GP

The answer is $\text{5}\text{.2}\text{cm}$ .

Explanation of Solution

Given:

The point charges are $\text{3}\text{.4μC}$ and $\text{-2}\text{.6μC}$ respectively and are placed $\text{1}\text{.6}\text{m}$ apart.

Formula used:

The magnitude of electric potential $\text{V}$ due to a point charge $\text{q}$ at a distance $\text{r}$ from the charge is given as

  $\text{V=k}\frac{\text{q}}{\text{r}}$

Where, $\text{k}$ is the Coulomb’s constant.

Calculation:

The electric potential is zero at such points where the electric potential subtended by the two charges cancel each other.

  ${\text{V}}_{\text{1}}{\text{=V}}_{\text{2}}$

  $\text{k}\frac{{\text{q}}_{\text{1}}}{\left(\text{d + r}\right)}\text{=k}\frac{{\text{q}}_{\text{2}}}{\left(\text{d + r}\right)}$

  $\frac{\text{r}}{\text{d + r}}\text{=}\frac{{\text{q}}_{\text{2}}}{{\text{q}}_{\text{1}}}$

  $\frac{\text{r}}{\text{1}\text{.6 + r}}\text{=}\frac{\text{2}{\text{.6 × 10}}^{\text{-6}}}{\text{3}{\text{.4 × 10}}^{\text{-6}}}$

  $\frac{\text{r}}{\text{1}\text{.6 + r}}\text{= 0}\text{.7647}$

  $\text{1}\text{.22352 + 0}\text{.7647r = r}$

  $\text{1}\text{.22352 = 0}\text{.2353 r}$

  $\text{r = 5}\text{.2}\text{cm}$