Question

2 particles having charges q1=0.500 nC and q2=8.00 nC are separated by a distance of 1.20m. at what point along the line connecting the two charges is the total electric field due to the two chares equal to zero?

Answer 1

Refer to the figure shown below.

Charge q₁ = 0.5 nC = 0.5x10⁻⁹ C
Charge q₂ = 8 nC = 8x10⁻⁹ C
d = 1.2 m, the distance between the two charges.

x is the distance between the two charges, measured from the charge q₁.

From Coulomb's Law,
The electric field generated along x by q₁ is
E₁ = k(q₁/x²)
The electric field generated along x by q₂ is
E₂ = -k[q₂/(d-x)²]
where
k = 8.988x10⁹ (N-m²)/C² is the Coulomb constant/

When the electric field along x is zero, then
E₁ + E₂ = 0
k[q₁/x² - q₂/(d-x)²] = 0

That is,
0.5/x² = 8/(1.2 - x)²
8x² = 0.5(1.2 - x)²
16x² = 1.44 - 2.4x + x²
15x² + 2.4x - 1.44 = 0

Solve with the quadratic formula.
x = (1/30)*[-2.4 +/- √(5.76 + 86.4)]
   = 0.24 or -0.4 m

Reject the negative answer to obtain
x = 0.24 m
d-x = 0.96 m

Answer
The electric field is zero between the charges so that
(a) It is at 0.24 m from the 0.5 nC charge, and
(b) It is at 0.96 m from the 8 nC charge.

Answer 2

The location of the point is at the distance of $\boxed{0.24{\text{ m}}}$from first particle and $\boxed{0.96{\text{ m}}}$ from second particle.

Further explanation:

Here, we have to calculate the location of the point on the line which connects the both charges at which the net electric field due to these charges is zero.

Given:

Charge on the first particle $\left( {{q_1}}\right)$ is $0.5{\text{nC}}=5\times {10^{- 10}}{\text{C}}$.

Charge on the second particle $\left({{q_2}}\right)$ is $8{\text{ nC}}=80\times{10^{-10}}{\text{C}}$.

Distance between the first particle and the second particle $\left(r\right)$ is $1.2{\text{ m}}$.

Formula and concept used:

Let us assume that the point is lying in between the charges as shown in Figure 1.

For, net electric field to be zero at that point,

The electric field due to first particle’s charge will be equal to the electric field due to second particle’s charge.

The expression can be written as,

$\boxed{{E_1}={E_2}}$

First of all we will know about the electric field.

Electric field: The electric field at a point due to a charge is define as the amount of force experienced by a unit charge also known as test charge at that point.

$E=\dfrac{1}{{4\pi{\varepsilon_0}}}\cdot \dfrac{q}{{{r^2}}}$

Here, $E$ is the electric field, $q$ is the charge and $r$ is the distance between charge and point at which electric field has to be measured.

Substitute the values of ${\vec E_1}$ and ${\vec E_2}$

We get,

$\dfrac{1}{{4\pi{\varepsilon_0}}} \cdot\dfrac{{{q_1}}}{{{x^2}}}=\dfrac{1}{{4\pi{\varepsilon_0}}} \cdot \dfrac{{{q_2}}}{{{{\left( {1.2 - x} \right)}^2}}}$

Simplify the above equation,

$\boxed{\dfrac{{{q_1}}}{{{{\left( x \right)}^2}}}=\dfrac{{{q_2}}}{{{{\left( {1.2 - x} \right)}^2}}}}$                              …… (1)

Calculation:

Substitute $5 \times {10^{- 10}}{\text{ C}}$ for ${q_1}$ and $80 \times {10^{ - 10}}{\text{ C}}$ for ${q_2}$ in equation (1).

$\begin{aligned}\frac{{5 \times {{10}^{- 10}}}}{{{{\left( x \right)}^2}}}&=\frac{{80\times {{10}^{- 10}}}}{{{{\left( {1.2 - x} \right)}^2}}}\\\frac{5}{{{x^2}}}&=\frac{{80}}{{{{\left( {1.2 - x} \right)}^2}}}\\5{\left( {1.2 - x}\right)^2}&=80{x^2}\\\end{aligned}$

Simplify the above equation,

$75{x^2} + 12x - 7.2 = 0$

Solve the above equation,  

$\begin{gathered}x=\frac{{-12 \pm \sqrt {{{12}^2} - 4\times75\times\left({-7.2} \right)}}}{{2 \times 75}}\\=\frac{{-12 \pm \sqrt {2304} }}{{150}}\\=\frac{{-12\pm48}}{{150}}\\\end{gathered}$

Taking positive value,

$\begin{aligned}x&=\frac{{ - 12 + 48}}{{150}}\\&=\frac{{36}}{{150}}\\&=0.24{\text{ m}}\\\end{aligned}$

Distance of the point from the second particle will be,

$\begin{aligned}\left({1.2-x}\right)&=1.2-0.24\\&=0.96{\text{ m}}\\\end{aligned}$

Thus, the location of the point is at the distance of $\boxed{0.24{\text{ m}}}$from first particle and $\boxed{0.96{\text{ m}}}$ from second particle.

Learn more:

1. Momentum change due to collision: brainly.com/question/9484203.  

2. Expansion of gas due to change in temperature: brainly.com/question/9979757.  

3. Conservation of momentum brainly.com/question/4033012.

Answer details:

Grade: Senior School  

Subject: Physics  

Chapter: Electric charges and fields

Keywords:

Two particles, separated by distance, point along line, zero electric field, coulomb law, charges, electric field, position vector, location of charge, 0.96m, 1m.