Question

In the figure below, determine the point (other than infinity) at which the total electric field is zero.
How many meters to the left of -2.5 × 10^-6 C charge

Answer 1

Concepts and reason

The concepts required to solve this problem is the electric field.

Initially, calculate the electric field at point P which lies on the x-axis, due charge${Q_1}$. Then, find the electric field at point P due charge${Q_2}$. Later, find the net electric field. The electric field at point P is zero. Finally, substitute the zero value for the electric field and solve the equation for x.

Fundamentals

The expression of the electric field is,

$E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{R^2}}}$

Here, E is the electric field, ${\varepsilon _0}$ is the permittivity of the free space, Q is the charge, and R is the distance.

Refer the figure given in the question.

Consider a point P on the x-axis which is x distance away from the ${Q_1}$ and $\left( {x + 1{\rm{ m}}} \right)$ away from the charge${Q_2}$.

The electric field ${E_1}$ at point P due charge${Q_1}$ is,

${E_1} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{{x^2}}}$

The electric field ${E_2}$ at point P due charge${Q_2}$ is,

${E_2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_2}}}{{{{\left( {x + 1{\rm{ m}}} \right)}^2}}}$

The electric field at point P is,

$E = {E_1} + {E_2}$

Substitute $\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{{x^2}}}$ for ${E_1}$ and $\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_2}}}{{{{\left( {x + 1{\rm{ m}}} \right)}^2}}}$ for${E_2}$.

$E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{{x^2}}} + \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_2}}}{{{{\left( {x + 1{\rm{ m}}} \right)}^2}}}$

Substitute 0 for E.

$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{{x^2}}} + \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_2}}}{{{{\left( {x + 1{\rm{ m}}} \right)}^2}}} = 0$

The electric field at point P is,

$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{{x^2}}} + \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_2}}}{{{{\left( {x + 1{\rm{ m}}} \right)}^2}}} = 0$

Substitute $- 2.5{\rm{ }}\mu {\rm{C}}$ for ${Q_1}$ and $6.0{\rm{ }}\mu {\rm{C}}$ for${Q_2}$.

$\begin{array}{c}\\\frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( { - 2.5{\rm{ }}\mu {\rm{C}}} \right)}}{{{x^2}}} + \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( {6.0{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {x + 1{\rm{ m}}} \right)}^2}}} = 0\\\\\frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( {2.5{\rm{ }}\mu {\rm{C}}} \right)}}{{{x^2}}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( {6.0{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {x + 1{\rm{ m}}} \right)}^2}}}\\\\\frac{{{x^2}}}{{{{\left( {x + 1{\rm{ m}}} \right)}^2}}} = \frac{{\left( {2.5{\rm{ }}\mu {\rm{C}}} \right)}}{{\left( {6.0{\rm{ }}\mu {\rm{C}}} \right)}}\\\end{array}$

Solve for x.

$\begin{array}{c}\\\frac{{{x^2}}}{{{{\left( {x + 1{\rm{ m}}} \right)}^2}}} = 0.4167\\\\{x^2} = 0.4167{\left( {x + 1{\rm{ m}}} \right)^2}\\\\0.5833{x^2} - 0.8334x - 0.4167 = 0\\\end{array}$

Solve the above quadratic equation.

$\begin{array}{c}\\{x_1} = 1.82{\rm{ m }}\\\\{x_2} = - 0.392{\rm{ m}}\\\end{array}$

The negative answer is not possible because in this case the point P will lie between the charges. So, the possible distance to the left of $- 2.5{\rm{ }}\mu {\rm{C}}$ charge is 1.82 m.

Ans:

The distance of the charge $- 2.5{\rm{ }}\mu {\rm{C}}$ from the point P is$1.82{\rm{ m}}$.

Answer 2

the electric field due to -ve charge will be towards the charge

the electric field due to +ve charge will be away from the charge

so the net electric field will be zero to the left of 2.5 uC

let the electric field be zero at a distance be d from the 2.5uC charge

electric field due to 6 uC = electric field due to -2.5 uC

k*2.5*10^-6 / (d^2) = k * 6*10^6 / (1+d)^2

2.5 (1+d^2 + 2d ) = 6d^2

3.5d^2 -2.5d - 2.5 = 0

d = 2.5 +- sqrt ( 2.5^2 +4*2.5*3.5) / 7

d = 0.56 m

the position is 0.56 m to the left of -2.5 uC

Answer 3