Chapter 18, Problem 15PE

To determine

(a)

The total electric field at x = 1.00 cm.

Answer to Problem 15PE

The total electric field at x = 1.00 cm is $5.8\times {10}^3\frac{\text{N}}{\text{C}}$.

Explanation of Solution

Given:

  

Point charges located at 1.00 cm, 5.00 cm, 8.00 cm and 14.0 cm along the x -axis.

Formula used:

Electric field in point P distanced d_i from n charges left of P is given by:

  $E_P={\displaystyle \sum _{i=1}^{n}k\frac{q_i}{d^2{}_i}}$

Calculation:

Electric field in point P distanced d_i from n charges left of P is given by:

  $E_P={\displaystyle \sum _{i=1}^{n}k\frac{q_i}{d^2{}_i}}$

  $\begin{array}{c}E=E_1+E_2+E_3\\ =\left(9\times {10}^9\times {10}^{-9}\right)\left[\frac{5}{{0.04}^2}+3\times \frac{5}{{0.07}^2}+\frac{5}{{0.13}^2}\right]\\ =9\left[3125+3061.2245+295.85\right]\\ =5.8\times {10}^3\frac{N}{C}\end{array}$

Conclusion:

The total electric field at x = 1.00 cm is $5.8\times {10}^3\frac{N}{C}$.

To determine

(b)

Find the total electric field at x = 11.00 cm in given figure.

Answer to Problem 15PE

The total electric field at x = 11.00 cm is $203500\frac{\text{N}}{C}$.

Explanation of Solution

Given info:

  

Point charges located at 1.00, 5.00, 8.00 and 14.0 cm along the x -axis.

Formula used:

Electric field in point P distanced d_i from n charges left of P is given with

  $E_P={\displaystyle \sum _{i=1}^{n}k\frac{q_i}{d^2{}_i}}$

Calculation:

Electric field in point P distanced d_i from n charges left of P is given with

  $E_P={\displaystyle \sum _{i=1}^{n}k\frac{q_i}{d^2{}_i}}$

  $E_{11}={\displaystyle \sum _{i=1}^{n}k\frac{q_i}{{\left(11\text{ cm}-x_i\right)}^2}}$

  $=9\cdot {10}^9\frac{{\text{Nm}}^2}{{\text{C}}^2}\cdot 5\text{nC}\left(\frac{-2}{{\left(10\text{ cm}\right)}^2}+\frac{1}{{\left(6\text{ cm}\right)}^2}+\frac{3}{{\left(\text{3 cm}\right)}^2}+\frac{1}{{\left(\text{3 cm}\right)}^2}\right)$

  $=203500\frac{\text{N}}{\text{C}}$

Note that the rightmost charge produces electric field rightwards at x = 11 cm.

Conclusion:

Thus, the total electric field at x = 11.00 cm is $203500\frac{\text{N}}{C}$.

To determine

(c)

Find the final charge configuration if the charges are allowed to move and eventually be brought to rest by friction.

Answer to Problem 15PE

The final charge configuration would be $+q\left(4\text{ charges}\right)$.

Explanation of Solution

Given info:

  

Point charges located at 1.00, 5.00, 8.00 and 14.0 cm along the x -axis.

Formula used:

Electric field in point P distanced d_i from n charges left of P is given with

  $E_P={\displaystyle \sum _{i=1}^{n}k\frac{q_i}{d^2{}_i}}$

Calculation:

Since + q and + 3q repel while − 2q and + q and + 3q and − q attract, - 2q will collide with + q forming a ball of charge − q and + 3q will collide with − q forming a ball of charge + 2q. Then, these two will collide forming a ball of charge + q somewhere along x axis. This ball of charge is made of 4 initial charges.

Conclusion:

The final charge configuration would be $+q\left(4\text{ charges}\right)$.