Question

4.) In the figure below (q₁ =-2.8 µC, q₂ =6.80 µC), determine the point (otherthan infinity) at which the electric field is zero.
m

2

between the two charges to theright of q₂    to theleft of q₁

Answer 1

Concepts and reason

The concepts used to solve this problem are the electric field and the quadratic equation.

Initially, draw the diagram of the electric fields due to the given charges for the given condition. Next, use the expression for the electric field to calculate the net electric field due to the given charges. Finally, use the expression for the roots of a quadratic equation to calculate the distance for the given condition.

Fundamentals

Electric field is defined as the force per unit charge. It is expressed mathematically as,

$E = \frac{F}{q}$

Here, $F$ is the force experienced by the charge and $q$ is the charge.

The direction of electric field is radially outward from positive charge and radially inward from negative charge. Electric field a distance $r$ away from a point charge $q$ is expressed as,

$E = \frac{{kq}}{{{r^2}}}$

Here, $k$ is proportionality constant and is equivalent to $9 \times {10^9}\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}$ .

For a condition, when the electric field due the two given charges that are in the opposite direction at a point, and their magnitudes are the same, the net electric field is zero. It means,

${E_1} + {E_2} = 0$

Here, E represents the electric field and subscripts denote the points 1 and 2.

Write the quadratic equation.

$a{x^2} + bx + c = 0$

Here, $x$ is the root of the equation, and coefficients $a$ , $b$ , and $c$ are the constants.

The roots of the quadratic equation are calculated as follows,

$x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

(4)

Consider the desired point Q is at a distance x measured to the left of charge ${q_1}$ .

Draw the diagram of the electric fields due to the given charges.

Here, ${q_1}$ and ${q_2}$ are the charges placed at a distance $1.00\;{\rm{m}}$ apart. ${E_1}$ is electric field due to ${q_1}$ and ${E_2}$ is electric field due to ${q_2}$ , these electric fields are at a point Q that is at a distance of $x$ from the charge ${q_1}$ .

The expression for electric field due to the charge ${q_1}$ is,

${E_1} = \frac{{k{q_1}}}{{{r^2}}}$

Substitute, $x$ for $r$ .

${E_1} = \frac{{k{q_1}}}{{{x^2}}}$

The expression for electric field due to the charge ${q_2}$ ,

${E_2} = \frac{{k{q_2}}}{{{r^2}}}$

Substitute, $\left( {1 + x} \right)$ for $r$ .

${E_2} = \frac{{k{q_2}}}{{{{\left( {1 + x} \right)}^2}}}$

The net electric field is equal to zero at point Q. Thus,

${E_1} + {E_2} = 0$

Substitute, $\frac{{k{q_1}}}{{{x^2}}}$ for ${E_1}$ and $\frac{{k{q_2}}}{{{{\left( {1 + x} \right)}^2}}}$ for ${E_2}$ .

$\begin{array}{l}\\\frac{{k{q_1}}}{{{x^2}}} + \frac{{k{q_2}}}{{{{\left( {1 + x} \right)}^2}}} = 0\\\\\frac{{ - {q_1}}}{{{x^2}}} = \frac{{{q_2}}}{{{{\left( {1 + x} \right)}^2}}}\\\end{array}$

Substitute, $- 2.80{\rm{ }}\mu {\rm{C}}$ for ${q_1}$ and $6.80{\rm{ }}\mu {\rm{C}}$ for ${q_2}$ .

$\begin{array}{l}\\\frac{{ - \left( { - 2.8\mu {\rm{C}}} \right)}}{{{x^2}}} = \frac{{\left( {6.80\mu {\rm{C}}} \right)}}{{{{\left( {1 + x} \right)}^2}}}\\\\\frac{{2.8}}{{{x^2}}} = \frac{{6.80}}{{{{\left( {1 + x} \right)}^2}}}\\\\2.8{\left( {1 + x} \right)^2} - 6.80{x^2} = 0\\\end{array}$

Further solve.

$- 4{x^2} + 5.6x + 2.8 = 0$

It is a quadratic equation. Solve the equation for the roots of the equation and to get the value of $x$ .

$a{x^2} + bx + c = 0$

The roots of the quadratic equation are calculated as follows,

$x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Substitute, $- 4$ for $a$ , $+ 5.6$ for $b$ and $+ 2.8$ for $c$ in the above expression of $x$

$\begin{array}{c}\\x = \frac{{ - \left( {5.6} \right) \pm \sqrt {{{\left( {5.6} \right)}^2} - \left( 4 \right)\left( { - 4} \right)\left( {2.8} \right)} }}{{\left( 2 \right)\left( { - 4} \right)}}\\\\ = \frac{{ - \left( {5.6} \right) \pm 8.727}}{{ - 8}}\\\\ = \frac{{ - \left( {5.6} \right) + 8.727}}{{ - 8}},\frac{{ - \left( {5.6} \right) - 8.727}}{{ - 8}}\\\\ = - 0.39{\rm{ m}},{\rm{ 1}}{\rm{.79 m}}\\\end{array}$

The negative value for $x$ is invalid because it locates the point between the charges where both fields are in same direction. Hence, take the positive value of $x$ .

Ans: Part 4

The point at which the net electric field is zero due to charges ${q_1}$ and ${q_2}$ is ${\rm{1}}{\rm{.79 m}}$ .