Determining Electrostatic Equilibrium for Charges on an Axis

This explanation details how to find the necessary value of a charge, denoted as q, to achieve a state of electrostatic equilibrium within a system of three charges positioned along the x-axis.

Understanding the System Setup

The problem involves three point charges placed linearly on the x-axis at specific locations:

• A charge of value $+Q$ is located at the origin, which is position $x=0$.
• A charge of value $q$ is located at position $x=d/2$.
• Another charge of value $+Q$ is located at position $x=d$.

Our objective is to calculate the value of the charge q that ensures the entire system remains in electrostatic equilibrium.

The Principle of Electrostatic Equilibrium

The fundamental condition for electrostatic equilibrium is that the net force acting on every single charge within the system must be exactly zero. If even one charge experiences a net force, the system is not in equilibrium. We will apply Coulomb's Law to calculate these forces.

Coulomb's Law quantifies the electrostatic force ($F$) between two point charges ($q_1$ and $q_2$) separated by a distance ($r$). It is expressed as:

$ F = \frac{k |q_1 q_2|}{r^2} $

Here, $k$ represents Coulomb's constant. The force acts along the line joining the two charges; it's repulsive if the charges have the same sign and attractive if they have opposite signs.

Analyzing the Forces on Each Charge

1. Force Analysis for $+Q$ at $x=0$

Let's consider the charge $+Q$ situated at the origin ($x=0$). This charge, let's call it $Q_1$, is influenced by the charge $q$ ($Q_2$) at $x=d/2$ and the charge $+Q$ ($Q_3$) at $x=d$.

• Interaction with $q$ ($Q_2$): The distance between $Q_1$ and $Q_2$ is $r_{12} = d/2$. The force exerted by $Q_2$ on $Q_1$, denoted $\vec{F}_{12}$, acts along the x-axis. If $q$ is positive, it's repulsive (positive x-direction); if $q$ is negative, it's attractive (negative x-direction). Using vector notation: $ \vec{F}_{12} = \frac{k Q q}{(d/2)^2} \hat{i} $ where $\hat{i}$ is the unit vector along the positive x-axis.
• Interaction with $+Q$ ($Q_3$): The distance between $Q_1$ and $Q_3$ is $r_{13} = d$. Since both are positive charges, the force $\vec{F}_{13}$ exerted by $Q_3$ on $Q_1$ is repulsive and points in the positive x-direction. $ \vec{F}_{13} = \frac{k Q Q}{d^2} \hat{i} $

For $Q_1$ to be in equilibrium, the vector sum of these forces must be zero:

$ \vec{F}_{net, 1} = \vec{F}_{12} + \vec{F}_{13} = 0 $

Substituting the expressions for the forces:

$ \left( \frac{k Q q}{(d/2)^2} + \frac{k Q Q}{d^2} \right) \hat{i} = 0 $

For this sum to be zero, the term in the parenthesis must be zero (assuming $k, Q, d$ are non-zero):

$ \frac{q}{(d/2)^2} + \frac{Q}{d^2} = 0 $

Let's simplify the term $(d/2)^2 = d^2/4$:

$ \frac{q}{d^2/4} + \frac{Q}{d^2} = 0 $

$ \frac{4q}{d^2} + \frac{Q}{d^2} = 0 $

Multiply the entire equation by $d^2$:

$ 4q + Q = 0 $

Solving for $q$:

$ q = -\frac{Q}{4} $

This result implies that the charge $q$ must be negative to balance the forces.

2. Force Analysis for $q$ at $x=d/2$

Now, let's examine the charge $q$ ($Q_2$) located at $x=d/2$. It is influenced by $Q_1$ ($+Q$) at $x=0$ and $Q_3$ ($+Q$) at $x=d$.

• Interaction with $+Q$ ($Q_1$): The distance is $r_{21} = d/2$. The force $\vec{F}_{21}$ exerted by $Q_1$ on $Q_2$ is directed towards $Q_1$ (negative x-direction). $ \vec{F}_{21} = \frac{k q Q}{(d/2)^2} (-\hat{i}) = -\frac{k q Q}{(d/2)^2} \hat{i} $
• Interaction with $+Q$ ($Q_3$): The distance is $r_{23} = d - d/2 = d/2$. The force $\vec{F}_{23}$ exerted by $Q_3$ on $Q_2$ is directed towards $Q_3$ (positive x-direction). $ \vec{F}_{23} = \frac{k q Q}{(d/2)^2} \hat{i} $

The net force on $Q_2$ is:

$ \vec{F}_{net, 2} = \vec{F}_{21} + \vec{F}_{23} = 0 $

$ \left( -\frac{k q Q}{(d/2)^2} + \frac{k q Q}{(d/2)^2} \right) \hat{i} = 0 $

$ 0 \cdot \hat{i} = 0 $

This equation $0=0$ is always true. This means the charge $q$ at the midpoint is always in equilibrium, provided $Q_1$ and $Q_3$ have the same sign and magnitude, because the forces they exert on $q$ are equal in magnitude and opposite in direction. This result does not constrain the value of $q$.

3. Force Analysis for $+Q$ at $x=d$

Finally, let's consider the charge $+Q$ ($Q_3$) at $x=d$. It is affected by $Q_1$ ($+Q$) at $x=0$ and $Q_2$ ($q$) at $x=d/2$.

• Interaction with $+Q$ ($Q_1$): The distance is $r_{31} = d$. The force $\vec{F}_{31}$ exerted by $Q_1$ on $Q_3$ is repulsive and acts towards $Q_1$ (negative x-direction). $ \vec{F}_{31} = \frac{k Q Q}{d^2} (-\hat{i}) = -\frac{k Q^2}{d^2} \hat{i} $
• Interaction with $q$ ($Q_2$): The distance is $r_{32} = d - d/2 = d/2$. The force $\vec{F}_{32}$ exerted by $Q_2$ on $Q_3$ depends on the sign of $q$. If $q$ is negative, this force is attractive and points towards $Q_2$ (negative x-direction). $ \vec{F}_{32} = \frac{k Q q}{(d/2)^2} (-\hat{i}) $

For $Q_3$ to be in equilibrium, the net force must be zero:

$ \vec{F}_{net, 3} = \vec{F}_{31} + \vec{F}_{32} = 0 $

$ \left( -\frac{k Q^2}{d^2} - \frac{k Q q}{(d/2)^2} \right) \hat{i} = 0 $

Setting the term in the parenthesis to zero (assuming $k, Q, d$ are non-zero):

$ -\frac{Q}{d^2} - \frac{q}{(d/2)^2} = 0 $

Substitute $(d/2)^2 = d^2/4$:

$ -\frac{Q}{d^2} - \frac{q}{d^2/4} = 0 $

$ -\frac{Q}{d^2} - \frac{4q}{d^2} = 0 $

Multiply by $-d^2$:

$ Q + 4q = 0 $

Solving for $q$:

$ 4q = -Q $

$ q = -\frac{Q}{4} $

Synthesizing the Results for System Equilibrium

For the entire system to achieve electrostatic equilibrium, the net force on each of the three charges must be zero. We found:

• The force on $Q_1$ ($+Q$ at $x=0$) requires $q = -Q/4$.
• The force on $Q_2$ ($q$ at $x=d/2$) is always zero due to symmetry.
• The force on $Q_3$ ($+Q$ at $x=d$) requires $q = -Q/4$.

Since both outer charges require $q = -Q/4$ for equilibrium, this is the value needed for the whole system. The negative sign confirms that $q$ must be an attractive charge relative to the positive charges to ensure equilibrium.

Matching the Result to the Options

The calculated value for the charge $q$ is $q = -Q/4$. Let's compare this with the given multiple-choice options:

• Option 1: $-Q/4$
• Option 2: $+Q/2$
• Option 3: $+Q/4$
• Option 4: $-Q/2$

Our derived value, $q = -Q/4$, corresponds exactly to Option 1.