Two fixed, equal, positive charges, each of magnitude q = 5 x 10^-5 C are located at points A and B

Q: Why is the kinetic energy at D zero?

Point D is defined as the farthest point the charge reaches. To turn back, the charge must momentarily come to a complete stop, meaning its velocity and kinetic energy drop to zero at that exact instant.

Q: Why do we multiply the potential energy by 2?

The moving charge is on the perpendicular bisector, meaning it is always equidistant from both fixed charges A and B. Since potential energy is a scalar quantity, the total potential energy is simply the sum of the identical potential energies from each fixed charge.

Q: Is the motion of the charge simple harmonic?

No. For motion to be simple harmonic, the restoring force must be directly proportional to the displacement ($F \propto -x$). Here, the force has a complex dependence on $x$ involving a denominator of $(x^2 + a^2)^{3/2}$. It is oscillatory, but not simple harmonic.

Q: What happens if the initial kinetic energy at C is greater than 9 J?

If the kinetic energy is greater than 9 J, the total mechanical energy of the system becomes positive. This means the charge has enough energy to completely escape the attractive potential well of the fixed charges and will travel to infinity without ever returning.

Q: Why is the potential energy negative?

The potential energy is negative because the electrostatic force between the positive fixed charges and the negative moving charge is attractive. A negative total energy indicates a bound system where the particle cannot escape to infinity.

Question

Two fixed, equal, positive charges, each of magnitude $q = 5 	imes 10^{-5}	ext{ C}$ are located at points $A$ and $B$ separated by a distance of 6 m. An equal and opposite charge moves towards them along the line $COD$, the perpendicular bisector of the line $AB$. The moving charge, when reaches the point $C$ at a distance of 4 m from $O$, has a kinetic energy of 4 J. Calculate the distance of the farthest point $D$ which the negative charge will reach before returning towards $C$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a tug-of-war, but with invisible electrostatic forces! We have two positive charges, each of magnitude $q = 5 	imes 10^{-5}	ext{ C}$, firmly anchored at points $A$ and $B$. They are separated by a distance of $6	ext{ m}$, meaning each is $3	ext{ m}$ away from the origin $O$.

Now, a negative charge $-q$ is moving along the perpendicular bisector of the line segment $AB$. It starts at point $C$, which is $4	ext{ m}$ away from the origin, and it's moving towards the center with a kinetic energy of $4	ext{ J}$. Our mission is to find out how far it will go on the other side before it stops and turns back. Let's call this farthest point $D$.

### The Master Equation: Energy Conservation

Since the only forces doing work here are conservative electrostatic forces, the total mechanical energy of the moving charge remains constant throughout its journey.

This means the sum of its kinetic and potential energy at point $C$ will be exactly equal to the sum of its kinetic and potential energy at the farthest point $D$. We can write this mathematically as:

$$K_C + U_C = K_D + U_D$$

### Calculating Energy at Point C

Let's figure out the total energy at point $C$. We already know the kinetic energy $K_C = 4	ext{ J}$. To find the potential energy $U_C$, we need the distance from the fixed charges to point $C$.

Using the Pythagorean theorem in the right-angled triangle $AOC$, the distance $AC$ is:

$$AC = \sqrt{OA^2 + OC^2} = \sqrt{3^2 + 4^2} = 5	ext{ m}$$

Because the setup is perfectly symmetric, the potential energy due to charge $A$ is the same as the potential energy due to charge $B$. So, the total potential energy at $C$ is simply twice the potential energy from one charge:

$$U_C = 2 \left[ \frac{1}{4\pi\varepsilon_0} \frac{q(-q)}{AC} ight]$$

Plugging in the values, where $\frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9	ext{ N m}^2/	ext{C}^2$:

$$U_C = 2 \left[ \frac{-9 	imes 10^9 	imes (5 	imes 10^{-5})^2}{5} ight] = -9	ext{ J}$$

So, the total mechanical energy of the system is $K_C + U_C = 4 - 9 = -5	ext{ J}$.

### The Farthest Point D

Now, let's look at point $D$. Because $D$ is the farthest point the charge reaches, it must momentarily come to a halt before reversing its direction. This means its kinetic energy at $D$ is zero ($K_D = 0$).

The potential energy at $D$ will depend on the unknown distance $AD$. Using the same logic as before:

$$U_D = 2 \left[ \frac{1}{4\pi\varepsilon_0} \frac{q(-q)}{AD} ight] = \frac{-45}{AD}$$

### Final Calculation

Now, we bring it all together using our energy conservation equation:

$$4 - 9 = 0 - \frac{45}{AD}$$

$$-5 = -\frac{45}{AD}$$

Solving for $AD$, we get:

$$AD = 9	ext{ m}$$

But we need the distance from the origin $O$ to $D$, which is $OD$. We can use the Pythagorean theorem one last time in triangle $AOD$:

$$OD = \sqrt{AD^2 - OA^2}$$

$$OD = \sqrt{9^2 - 3^2} = \sqrt{81 - 9} = \sqrt{72}$$

$$OD \approx 8.48	ext{ m}$$

**The negative charge will reach a maximum distance of 8.48 m from the center before returning.**

Enter Numerical Value:

Visualized Solution

$S e t u p o f t h e c ha r g es$

$K_{C} + U_{C} = K_{D} + U_{D}$

$A C = O A^{2} + O C^{2}$

$U_{C} = - 9 J$

$K_{D} = 0$

$A D = 9 m$

$O D \approx 8.48 m$

$I s t h e m o t i o n S im pl e H a r m o ni c ?$

The Sigma Insight: Electrostatic Potential Energy

Analyzing the Setup

The Master Equation

Calculating Energy at Point C

The Farthest Point D

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setup of the charges

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } KC​+UC​=KD​+UD​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC=OA2+OC2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } UC​=−9 J

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } KD​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AD=9 m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OD≈8.48 m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Is the motion Simple Harmonic?