Chapter 16, Problem 35P

To determine

To find: The magnitude and direction of field

Answer to Problem 35P

  $\frac{-4\mathrm{k}\mathrm{Q}\mathrm{a}\mathrm{x}}{{\left({\mathrm{x}}^2-{\mathrm{a}}^2\right)}^2}$towards left.

Explanation of Solution

Given:

Separation between charges $=2\mathrm{a}$

Formula used:

Expression for the electric field at point $\text{P}$ due to charge $\mathrm{Q}$ is,

  $\mathrm{E}=\frac{\mathrm{k}\mathrm{Q}}{{\mathrm{x}}^2}$

Here, E is the electric field and x is distance between the charges.

Calculation:

  

Choose the right direction to be positive. Then the field due to + Q will be positive, and the field

due to - Q will be negative.

The electric field at P due to the charge - Q is,

  ${\mathrm{E}}_1=-\frac{\mathrm{k}\mathrm{Q}}{{(\mathrm{x}-\mathrm{a})}^2}\dots \dots (1)$

The electric field at P due to the charge -Q is,

  ${\mathrm{E}}_2=\frac{\mathrm{k}\mathrm{Q}}{{(\mathrm{x}+\mathrm{a})}^2}\dots \dots (2)$

Hence, the magnitude of net electric field at P is expressed as follows:

  $\mathrm{E}={\mathrm{E}}_1+{\mathrm{E}}_2$

Substitute $\frac{\mathrm{k}\mathrm{Q}}{{(\mathrm{x}-\mathrm{a})}^2}$ for ${\mathrm{E}}_1$ and $\frac{\mathrm{k}\mathrm{Q}}{{(\mathrm{x}+\mathrm{a})}^2}$ for ${\mathrm{E}}_2$

  $\begin{array}{l}\mathrm{E}={\mathrm{E}}_1+{\mathrm{E}}_2\\ \begin{array}{l}=-\frac{\mathrm{k}\mathrm{Q}}{{(\mathrm{x}-\mathrm{a})}^2}+\frac{\mathrm{k}\mathrm{Q}}{{(\mathrm{x}+\mathrm{a})}^2}\hfill \\ =\mathrm{k}\mathrm{Q}\left[\frac{1}{{(\mathrm{x}+\mathrm{a})}^2}-\frac{1}{{(\mathrm{x}-\mathrm{a})}^2}\right]\hfill \\ =\mathrm{k}\mathrm{Q}\left[\frac{\left({\mathrm{x}}^2+{\mathrm{a}}^2-2\mathrm{a}\mathrm{x}\right)-\left({\mathrm{x}}^2+{\mathrm{a}}^2+2\mathrm{a}\mathrm{x}\right)}{{\left({\mathrm{x}}^2-{\mathrm{a}}^2\right)}^2}\right]\hfill \end{array}\end{array}$

Conclusion:

Therefore, the required magnitude and direction of the electric field at point P is $\frac{-4\mathrm{k}\mathrm{Q}\mathrm{a}\mathrm{x}}{{\left({\mathrm{x}}^2-{\mathrm{a}}^2\right)}^2}$ and towards the left.