Question

Two fixed, equal, positive charges, each of magnitude $q = 5 \times 10^{-5}$ C are located at points $A$ and $B$ separated by a distance of 6 m as shown in figure. An equal and opposite charge moves towards them along the line $COD$, the perpendicular bisector of the line $AB$. The moving charge, when it reaches the point $C$ at a distance of 4 m from $O$, has a kinetic energy of 4 J. Let $D$ be the farthest point where the negative charge will reach before returning towards $C$. Find distance $AD$ (in m)

Answer 1

9

Answer 2

The problem involves the conservation of mechanical energy for a charged particle moving in an electrostatic field.

1. Define the system and given parameters:

• Two fixed positive charges: $q_A = q_B = q = 5 \times 10^{-5}$ C.
• Distance between A and B: $AB = 6$ m. Since O is the midpoint, $OA = OB = 3$ m.
• Moving charge: $-q'$ (stated as "equal and opposite charge", so $q' = q = 5 \times 10^{-5}$ C).
• Coulomb's constant: $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.

2. Calculate the potential energy function:

Let the moving charge be at a point P on the perpendicular bisector at a distance $x$ from O.
The distance from A (or B) to P is $r = \sqrt{OA^2 + OP^2} = \sqrt{3^2 + x^2} = \sqrt{9 + x^2}$.
The electrostatic potential energy $U(x)$ of the charge $-q'$ at point P due to the two fixed charges $q$ is:

$U(x) = k \frac{q(-q')}{r} + k \frac{q(-q')}{r}$
$U(x) = -2k \frac{qq'}{r}$
$U(x) = -2k \frac{qq'}{\sqrt{9 + x^2}}$

Now, calculate the constant $2kqq'$:

$2kqq' = 2 \times (9 \times 10^9 \text{ N m}^2/\text{C}^2) \times (5 \times 10^{-5} \text{ C}) \times (5 \times 10^{-5} \text{ C})$
$2kqq' = 18 \times 10^9 \times 25 \times 10^{-10}$
$2kqq' = 450 \times 10^{-1} = 45$ J m.

So, the potential energy function is:

$U(x) = -\frac{45}{\sqrt{9 + x^2}}$ J.

3. Calculate the total mechanical energy at point C:

At point C, the distance from O is $OC = 4$ m.
The kinetic energy at C is $K_C = 4$ J.
The potential energy at C is $U_C = U(x=4)$:

$U_C = -\frac{45}{\sqrt{9 + 4^2}} = -\frac{45}{\sqrt{9 + 16}} = -\frac{45}{\sqrt{25}} = -\frac{45}{5} = -9$ J.
The total mechanical energy $E$ is the sum of kinetic and potential energy:

$E = K_C + U_C = 4 \text{ J} + (-9 \text{ J}) = -5$ J.

4. Apply conservation of energy to find the position of point D:

Point D is the farthest point the negative charge reaches, meaning its kinetic energy at D is zero ($K_D = 0$). Due to conservation of mechanical energy, the total energy at D must be equal to the total energy at C:

$E = K_D + U_D$
$-5 \text{ J} = 0 \text{ J} + U_D$
So, $U_D = -5$ J.

Let the distance of point D from O be $OD = x_D$.
Using the potential energy function:

$U_D = -\frac{45}{\sqrt{9 + x_D^2}}$
$-5 = -\frac{45}{\sqrt{9 + x_D^2}}$
$5 = \frac{45}{\sqrt{9 + x_D^2}}$
$\sqrt{9 + x_D^2} = \frac{45}{5}$
$\sqrt{9 + x_D^2} = 9$

Square both sides:

$9 + x_D^2 = 9^2$
$9 + x_D^2 = 81$
$x_D^2 = 81 - 9$
$x_D^2 = 72$
$x_D = \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$ m.
This is the distance $OD$.

5. Calculate the distance AD:

Point A is at a distance of 3 m from O. Point D is at a distance $x_D = 6\sqrt{2}$ m from O along the perpendicular bisector.
Using the Pythagorean theorem in $\triangle AOD$:

$AD = \sqrt{OA^2 + OD^2}$
$AD = \sqrt{3^2 + (6\sqrt{2})^2}$
$AD = \sqrt{9 + 72}$
$AD = \sqrt{81}$
$AD = 9$ m.

The final answer is $\boxed{9}$ m.