Chapter 18, Problem 16PE

To determine

(a)

The electric field at x = 5.00 cm.

Answer to Problem 16PE

The electric field at x = 5.00 cm is $4\times {10}^7\frac{\text{N}}{C}$.

Explanation of Solution

Given:

Point charges located at 3.00, 8.00 and 11.0 cm along the x -axis.

Formula used:

Electric field in point P distanced d_i from n charges on a line is given with

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

Where positive sign is taken for charges left to the point P and negative for charges right to P.

Calculation:

  $E_{\left(x=5\right)}={\displaystyle \sum _{i=1}^{3}k\frac{\pm q_i}{{\left(x_i-5\text{ cm}\right)}^2}}$

  $=9\times {10}^9\frac{{\text{Nm}}^2}{{\text{C}}^2}\times {10}^{-6}\text{C}\times \left(\frac{1}{{0.02}^2}-\frac{-2}{{0.03}^2}-\frac{1}{{0.06}^2}\right)\frac{1}{{\text{m}}^2}$

  $=4\cdot {10}^7\frac{\text{N}}{C}$

Conclusion:

The electric field at x = 5.00 cm is $4\times {10}^7\frac{\text{N}}{C}$.

To determine

(b)

To Find:

The position between 3.00 and 8.00 cm where the total electric field the same as that for − 2q alone.

Answer to Problem 16PE

The position is 7 cm.

Explanation of Solution

Given:

Point charges located at 3.00, 8.00 and 11.0 cm along the x -axis.

Calculation:

The charges will cancel out at the mid-point of $x=3\text{ cm and }x=11\text{ cm}\text{.}$

So, we get

  $\begin{array}{c}\frac{x_3+x_{11}}{2}=\frac{3\text{ cm}+11\text{ cm}}{2}\\ =\frac{14\text{ cm}}{2}\\ =7\text{ cm}\end{array}$

Conclusion:

Thus, the position is at 7 cm.

To determine

(c)

Whether the electric field between 0.00 and 8.00 cmbe zero.

Answer to Problem 16PE

Electric filed will never be zero.

Explanation of Solution

Given:

  

Point charges located at 3.00, 8.00 and 11.0 cm along the x -axis.

Calculation:

Note that such cancellation can't happen between a $3<x<8$ because rightmost charge is too small to parry the other two in that area. One way of solving this is to set up and solve an equation,

  $\begin{array}{l}E=k\left(\frac{q\times {10}^{-4}}{{\left(3-x\right)}^2}+\frac{q\times {10}^{-4}}{{\left(11-x\right)}^2}-\frac{q\times {10}^{-4}}{{\left(8-x\right)}^2}\right)=0\\ \Rightarrow \frac{q\times {10}^{-4}}{{\left(3-x\right)}^2}+\frac{q\times {10}^{-4}}{{\left(11-x\right)}^2}-\frac{q\times {10}^{-4}}{{\left(8-x\right)}^2}=0\\ \Rightarrow \frac{1}{{\left(3-x\right)}^2}+\frac{1}{{\left(11-x\right)}^2}=\frac{1}{{\left(8-x\right)}^2}\end{array}$

which is not possible.

In other words, when it reaches 0 after continuously dropping in some direction, electric field will change direction.

Since ${\text{lim}}_{x\to 3^{-}}E$ points leftwards, it's enough to check for x = 0:

  $E\left(0\right)=kq\left(\frac{1}{3^2}+\frac{-2}{8^2}+\frac{1}{{11}^2}\right)\frac{1}{{\text{cm}}^2}>0$

Which means it points to the left, so E = 0 can't happen neither it can happen between $0<x<3$.

Conclusion:

Electric field will never be zero.

To determine

(d)

To Find: where the electric field most rapidly approaches zero for very large positive or negative values of x in case (a) and (b).

Answer to Problem 16PE

Electric field of configuration (a) drops more rapidly.

Explanation of Solution

Given:

Point charges located at 3.00, 8.00 and 11.0 cm along the x -axis.

Calculation:

Total charge in (a) is zero, while total charge in (b) is not zero.

Looked from infinity, those charges may be considered emerging from the same point, so electric field in (a) will drop more rapidly.

Conclusion:

Thus, Electric field of configuration (a) drops more rapidly

To determine

(e)

Find the position to the right of 11.00 cm where the total electric field is zero, other than at infinity.

Answer to Problem 16PE

The position to the right of 11.00 cm where the electric field is zero other than infinity is 30.872 cm.

Explanation of Solution

Given info:

  

Point charges located at 3.00, 8.00 and 11.0 cm along the x -axis.

Calculation:

After cancelling some constants, we can obtain an equation,

  $\frac{1}{x^2}-\frac{2}{{\left(x+3\right)}^2}+\frac{1}{{\left(x+8\right)}^2}=0$

This after some algebraic manipulation converts to,

  $-2x^3+25x^2+264x+576=0$

With one real solution, x = 19.872 cm. since x denotes distance from the rightmost charge, coordinate of such point in the system given is,

  $x_0=11\text{ cm}+x=30.872\text{ cm}$

Conclusion:

Thus, the position to the right of 11.00 cm where the electric field is zero other than infinity is 30.872 cm.