Question

Two point charges +Q are placed on the x-axis at $x=+\frac{d}{2}$ and $x=-\frac{d}{2}$ as shown in the figure. A third charge $q$ is placed on the y-axis such that the force on the charge $q$ is maximum. The distance of charge $q$ from the origin is $\frac{d}{\sqrt{n}}$. Value of n, is ________.

Answer 1

8

Answer 2

To determine the distance from the origin where the force on charge $q$ is maximum, we follow these steps:

1. Define coordinates and forces:

   • The two charges $+Q$ are located at $P_1(-\frac{d}{2}, 0)$ and $P_2(+\frac{d}{2}, 0)$.
   • The charge $q$ is placed on the y-axis at $P_q(0, y)$.
   • The distance from each $+Q$ charge to $q$ is $r = \sqrt{(\frac{d}{2})^2 + y^2}$.

2. Calculate the force due to each $+Q$ charge:

   • The magnitude of the force exerted by each $+Q$ charge on $q$ is given by Coulomb's Law: $F_Q = k \frac{Q q}{r^2} = k \frac{Q q}{(\frac{d}{2})^2 + y^2}$ where $k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.

3. Resolve forces into components:

   • Let $\theta$ be the angle that the line connecting a $+Q$ charge to $q$ makes with the y-axis.
   • From the geometry, $\cos\theta = \frac{y}{r} = \frac{y}{\sqrt{(\frac{d}{2})^2 + y^2}}$.
   • Due to symmetry, the x-components of the forces from $P_1$ and $P_2$ on $q$ will cancel out.
   • The y-components will add up. The net force $F_{net}$ on $q$ will be along the y-axis.
   • $F_{net} = 2 F_Q \cos\theta$
   • Substitute the expressions for $F_Q$ and $\cos\theta$: $F_{net} = 2 \left( k \frac{Q q}{(\frac{d}{2})^2 + y^2} \right) \left( \frac{y}{\sqrt{(\frac{d}{2})^2 + y^2}} \right)$ $F_{net} = \frac{2 k Q q y}{((\frac{d}{2})^2 + y^2)^{3/2}}$

4. Find the condition for maximum force:

   • To find the value of $y$ for which $F_{net}$ is maximum, we differentiate $F_{net}$ with respect to $y$ and set the derivative to zero ($dF_{net}/dy = 0$).
   • Let $C = 2kQq$ and $a = \frac{d}{2}$. So, $F_{net} = C \frac{y}{(a^2 + y^2)^{3/2}}$.
   • Using the quotient rule or product rule for differentiation: $\frac{dF_{net}}{dy} = C \left[ \frac{(a^2 + y^2)^{3/2} \cdot 1 - y \cdot \frac{3}{2}(a^2 + y^2)^{1/2} \cdot (2y)}{((a^2 + y^2)^{3/2})^2} \right]$ $\frac{dF_{net}}{dy} = C \left[ \frac{(a^2 + y^2)^{3/2} - 3y^2(a^2 + y^2)^{1/2}}{(a^2 + y^2)^3} \right]$
   • For $F_{net}$ to be maximum, the numerator must be zero: $(a^2 + y^2)^{3/2} - 3y^2(a^2 + y^2)^{1/2} = 0$
   • Divide by $(a^2 + y^2)^{1/2}$ (since it's non-zero): $(a^2 + y^2) - 3y^2 = 0$ $a^2 - 2y^2 = 0$ $a^2 = 2y^2$ $y^2 = \frac{a^2}{2}$
   • Substitute $a = \frac{d}{2}$: $y^2 = \frac{(d/2)^2}{2} = \frac{d^2/4}{2} = \frac{d^2}{8}$
   • The distance of charge $q$ from the origin is $|y|$: $|y| = \sqrt{\frac{d^2}{8}} = \frac{d}{\sqrt{8}}$

5. Compare with the given format:

   • The problem states that the distance is $\frac{d}{\sqrt{n}}$.
   • Comparing our result $\frac{d}{\sqrt{8}}$ with $\frac{d}{\sqrt{n}}$, we get: $\sqrt{n} = \sqrt{8}$ $n = 8$