Question

Two equal point charges A and B are R distance apart. A third point charge placed on the perpendicular bisector at a distance 'x' from the centre will experience maximum electrostatic force when :-

• A. x = $\frac{R}{2\sqrt{2}}$
• B. x = $\frac{R}{\sqrt{2}}$
• C. x = R$\sqrt{2}$
• D. x = 2$\sqrt{2}$R

Answer 1

Answer: (A)

x = $\frac{R}{2\sqrt{2}}$

Answer 2

To determine the position 'x' where the third point charge experiences maximum electrostatic force, we follow these steps:

1. Set up the geometry and forces:
   Let the two equal point charges be $Q$ and the third point charge be $q$.
   The distance between charges A and B is $R$. Let the origin be the midpoint of A and B. So, charge A is at $(-R/2, 0)$ and charge B is at $(R/2, 0)$.
   The third charge $q$ is placed on the perpendicular bisector at $(0, x)$.
   The distance from charge A (or B) to charge $q$ is $r = \sqrt{(R/2)^2 + x^2} = \sqrt{R^2/4 + x^2}$.

2. Calculate the force due to one charge:
   The magnitude of the force exerted by charge A (or B) on charge $q$ is given by Coulomb's Law:
   $F = \frac{kQq}{r^2} = \frac{kQq}{R^2/4 + x^2}$, where $k = \frac{1}{4\pi\varepsilon_0}$.

3. Determine the net force:
   Due to symmetry, the horizontal components of the forces from A and B on $q$ cancel out. The vertical components add up.
   Let $\theta$ be the angle between the line connecting A to $q$ (or B to $q$) and the perpendicular bisector (y-axis).
   From the geometry, $\cos\theta = \frac{x}{r} = \frac{x}{\sqrt{R^2/4 + x^2}}$.
   The net force $F_{net}$ on charge $q$ is the sum of the vertical components:
   $F_{net} = F \cos\theta + F \cos\theta = 2F \cos\theta$
   Substitute the expressions for $F$ and $\cos\theta$:
   $F_{net} = 2 \left( \frac{kQq}{R^2/4 + x^2} \right) \left( \frac{x}{\sqrt{R^2/4 + x^2}} \right)$
   $F_{net} = \frac{2kQqx}{(R^2/4 + x^2)^{3/2}}$

4. Find the maximum force using differentiation:
   To find the value of $x$ for which $F_{net}$ is maximum, we differentiate $F_{net}$ with respect to $x$ and set the derivative to zero ($\frac{dF_{net}}{dx} = 0$).
   Let $C = 2kQq$ (a constant). So, $F_{net}(x) = C \frac{x}{(R^2/4 + x^2)^{3/2}}$.
   Using the quotient rule or product rule:
   $\frac{dF_{net}}{dx} = C \frac{d}{dx} \left[ x (R^2/4 + x^2)^{-3/2} \right]$
   $\frac{dF_{net}}{dx} = C \left[ 1 \cdot (R^2/4 + x^2)^{-3/2} + x \cdot \left(-\frac{3}{2}\right) (R^2/4 + x^2)^{-5/2} (2x) \right]$
   $\frac{dF_{net}}{dx} = C \left[ (R^2/4 + x^2)^{-3/2} - 3x^2 (R^2/4 + x^2)^{-5/2} \right]$
   Set $\frac{dF_{net}}{dx} = 0$:
   $(R^2/4 + x^2)^{-3/2} - 3x^2 (R^2/4 + x^2)^{-5/2} = 0$
   Factor out $(R^2/4 + x^2)^{-5/2}$:
   $(R^2/4 + x^2)^{-5/2} \left[ (R^2/4 + x^2)^{(-3/2) - (-5/2)} - 3x^2 \right] = 0$
   $(R^2/4 + x^2)^{-5/2} \left[ (R^2/4 + x^2)^1 - 3x^2 \right] = 0$
   Since $(R^2/4 + x^2)^{-5/2}$ cannot be zero (for real $x$), we must have:
   $R^2/4 + x^2 - 3x^2 = 0$
   $R^2/4 - 2x^2 = 0$
   $R^2/4 = 2x^2$
   $x^2 = \frac{R^2}{8}$
   $x = \sqrt{\frac{R^2}{8}} = \frac{R}{\sqrt{8}} = \frac{R}{2\sqrt{2}}$ (since x is a distance, it must be positive).

The force will be maximum when $x = \frac{R}{2\sqrt{2}}$.