Chapter 16, Problem 66E

(a)

To determine

The magnitude and direction of electric field at $x=1\text{m}$, $y=0\text{m}$.

Answer to Problem 66E

The magnitude of electric field at $x=1\text{m}$, $y=0\text{m}$ is $\underset{\_}{9.0\times {10}^3\text{N}\cdot {\text{C}}^{-1}}$ and it is directed towards negative $x$ axis..

Explanation of Solution

Given that the charge $-{10}^{-6}\text{C}$ is placed at the origin.

Write the expression for the of electric field at a point due to a point charge.

  $E=\frac{kQ}{r^2}$        (I)

Here, $E$ is the electric field, $Q$ is the charge, $r$ is the distance of the charge from the point, and $k$ is Coulomb constant ($k=9.0\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}$).

Conclusion:

Substitute $9.0\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}$ for $k$, $-{10}^{-6}\text{C}$ for $Q$, and $1.0\text{m}$ for $r$ in equation (I) to find $E$.

  $\begin{array}{c}E=\frac{\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}\right)\left(-1.0\times {10}^{-6}\text{C}\right)}{{\left(1.0\text{m}\right)}^2}\\ =-9.0\times {10}^3\text{N}\cdot {\text{C}}^{-1}\end{array}$

The magnitude of electric field is,

$\left|E\right|=9.0\times {10}^3\text{N}\cdot {\text{C}}^{-1}$

The direction of electric field is radially outward from a positive charge and radially inward to a negative charge.

Since, the given charge is negative; the electric field will be towards the charge. The direction of electric field is towards negative $x$ axis.

Therefore, the magnitude of electric field between the plates is $\underset{\_}{9.0\times {10}^3\text{N}\cdot {\text{C}}^{-1}}$ and it is directed towards negative $x$ axis.

(b)

To determine

The magnitude and direction of electric field at $x=0\text{m}$, $y=-2\text{m}$.

Answer to Problem 66E

The magnitude of electric field at $x=0\text{m}$, $y=-2\text{m}$ is $\underset{\_}{2.2\times {10}^3\text{N}\cdot {\text{C}}^{-1}}$ and it is directed towards positive $y$ axis.

Explanation of Solution

Given that the charge $-{10}^{-6}\text{C}$ is placed at the origin.

Write the expression for the electric field at a point due to a point charge.

  $E=\frac{kQ}{r^2}$        (II)

Here, $E$ is the electric field, $Q$ is the charge, $r$ is the distance of the charge from the point, and $k$ is Coulomb constant ($k=9.0\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}$).

Conclusion:

Substitute $9.0\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}$ for $k$, $-{10}^{-6}\text{C}$ for $Q$, and $-2.0\text{m}$ for $r$ in equation (II) to find $E$.

  $\begin{array}{c}E=\frac{\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}\right)\left(-1.0\times {10}^{-6}\text{C}\right)}{{\left(-2.0\text{m}\right)}^2}\\ =-2.2\times {10}^3\text{N}\cdot {\text{C}}^{-1}\end{array}$

The magnitude of electric field is,

$\left|E\right|=2.2\times {10}^3\text{N}\cdot {\text{C}}^{-1}$

The direction of electric field is radially outward from a positive charge and radially inward to a negative charge.

Since, the given charge is negative; the electric field will be towards the charge. The direction of electric field is towards positive $y$ axis.

Therefore, the magnitude of electric field at $x=0\text{m}$, $y=-2\text{m}$ is $\underset{\_}{2.2\times {10}^3\text{N}\cdot {\text{C}}^{-1}}$ and it is directed towards positive $y$ axis.

(c)

To determine

The magnitude and direction of electric field at $x=2\text{m}$, $y=2\text{m}$.

Answer to Problem 66E

The magnitude of electric field at $x=2\text{m}$, $y=2\text{m}$ is $\underset{\_}{1.1\times {10}^3\text{N}\cdot {\text{C}}^{-1}}$ and it is directed towards the charge from the point.

Explanation of Solution

Draw the figure for charge and the point.

From the figure, use Pythagoras theorem to calculate the distance of the point from the origin

Write the expression for the distance of the point from the charge,

$r=\sqrt{x^2+y^2}$

Here, $r$ is the distance of the point from the charge.

Substitute $2.0\text{m}$ for $x$ and $2.0\text{m}$ for $y$.

$\begin{array}{c}r=\sqrt{{\left(2.0\text{m}\right)}^2+{\left(2.0\text{m}\right)}^2}\\ =2.8\text{m}\end{array}$

Given that the charge $-{10}^{-6}\text{C}$ is placed at the origin.

Write the expression for the electric field at a point due to a point charge.

  $E=\frac{kQ}{r^2}$        (III)

Here, $E$ is the electric field, $Q$ is the charge, $r$ is the distance of the charge from the point, and $k$ is Coulomb constant ($k=9.0\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}$).

Conclusion:

Substitute $9.0\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}$ for $k$, $-{10}^{-6}\text{C}$ for $Q$, and $-2.0\text{m}$ for $r$ in equation (III) to find $E$.

  $\begin{array}{c}E=\frac{\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}\right)\left(-1.0\times {10}^{-6}\text{C}\right)}{{\left(2.8\text{m}\right)}^2}\\ =1.1\times {10}^3\text{N}\cdot {\text{C}}^{-1}\end{array}$

The magnitude of electric field is,

$\left|E\right|=1.1\times {10}^3\text{N}\cdot {\text{C}}^{-1}$

The direction of electric field is radially outward from a positive charge and radially inward to a negative charge.

Since, the given charge is negative; the electric field will be towards the charge from the point.

Therefore, the magnitude of electric field at $x=2\text{m}$, $y=2\text{m}$ is $\underset{\_}{1.1\times {10}^3\text{N}\cdot {\text{C}}^{-1}}$ and it is directed towards the charge from the point.