Four point charges +8 uC, -1 uC, -1 uC and +8 uC are fixed at the points -sqrt(27/2) m, -sqrt(3/2) m

Q: Why do we need to find the maximum potential?

The particle must overcome the highest potential barrier along its path to reach the origin. If it has enough energy to cross the peak, it will reach the origin.

Q: Why is the electric field zero at the maximum potential?

By definition, the electric field is the negative gradient of the potential ($E = -dV/dx$). At a local maximum or minimum of potential, the gradient is zero.

Q: Why does the particle speed up after crossing the peak?

After the peak, the potential decreases towards the origin. As potential energy decreases, kinetic energy increases due to conservation of energy.

Q: What happens if the initial speed is less than $3\text{ m/s}$?

The particle will not have enough kinetic energy to overcome the potential barrier. It will slow down, stop momentarily before the peak, and then be repelled back towards infinity.

Q: Why is the potential at the origin not the maximum?

The negative charges are closer to the origin than the positive charges. Their attractive potential creates a local 'dip' or well at the origin, making the peak potential occur at some distance away.

Question

Four point charges $+8\mu	ext{C}, -1\mu	ext{C}, -1\mu	ext{C}$ and $+8\mu	ext{C}$ are fixed at the points $-\sqrt{27/2}	ext{ m}, -\sqrt{3/2}	ext{ m}, +\sqrt{3/2}	ext{ m}$ and $+\sqrt{27/2}	ext{ m}$ respectively on the $y$-axis. A particle of mass $6 	imes 10^{-4}	ext{ kg}$ and charge $+0.1\mu	ext{C}$ moves along the $x$-direction. Its speed at $x = +\infty$ is $v_0$. Find the least value of $v_0$ for which the particle will cross the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity free. $(1/4\pi\varepsilon_0 = 9 	imes 10^9 	ext{ Nm}^2/	ext{C}^2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a particle embarking on a journey along the $x$-axis, starting from infinity and heading towards the origin. Along the $y$-axis, four charges are standing guard like sentinels. Two positive charges ($+8\mu	ext{C}$) are placed further out, while two negative charges ($-1\mu	ext{C}$) are positioned closer to the origin.

Our traveler is a positively charged particle ($+0.1\mu	ext{C}$). As it moves closer, it feels the pull of the negative charges and the push of the positive ones. Because the positive charges are much larger in magnitude, their repulsive effect will eventually dominate, creating a "potential mountain" that our particle must climb.

### The Potential Landscape
To understand this mountain, we need to map out the electric potential $V(x)$ along the $x$-axis. The total potential at any point $x$ is the sum of the potentials created by each of the four charges.

Using the formula for the potential of a point charge, $V = \frac{Kq}{r}$, we can write the total potential as:
$$V(x) = \frac{2KQ}{\sqrt{y_Q^2 + x^2}} - \frac{2Kq}{\sqrt{y_q^2 + x^2}}$$

Substituting the given values for the charges and their positions, we get a precise mathematical description of our landscape:
$$V(x) = 1.8 	imes 10^4 \left[ \frac{8}{\sqrt{\frac{27}{2} + x^2}} - \frac{1}{\sqrt{\frac{3}{2} + x^2}} ight]$$

### Finding the Peak of the Mountain
To find the minimum speed required to reach the origin, we must find the highest point of this potential mountain. At the peak, the slope of the potential is zero, which means the electric field $E = -\frac{dV}{dx}$ is zero.

By differentiating our potential function and setting it to zero, we find the exact location of the peak:
$$\frac{8x}{\left(\frac{27}{2} + x^2ight)^{3/2}} = \frac{x}{\left(\frac{3}{2} + x^2ight)^{3/2}}$$

Solving this algebraic puzzle reveals that the peak is located at $x = \sqrt{\frac{5}{2}}	ext{ m}$. Now, we plug this coordinate back into our potential equation to find the height of the mountain:
$$V_{	ext{max}} = 2.7 	imes 10^4	ext{ V}$$

### The Minimum Speed to Cross
Our particle starts at infinity, where the potential is zero. To just barely make it over the peak, its initial kinetic energy must exactly equal the potential energy at the top of the mountain. This is the principle of conservation of energy in action!

$$\frac{1}{2}mv_0^2 = q_0 V_{	ext{max}}$$

By plugging in the mass, charge, and maximum potential, we can solve for the critical initial speed:
$$v_0 = \sqrt{\frac{2 	imes 10^{-7} 	imes 2.7 	imes 10^4}{6 	imes 10^{-4}}} = 3	ext{ m/s}$$

If the particle is launched with a speed of exactly $3	ext{ m/s}$, it will coast to a halt right at the peak, and then gently fall down the other side towards the origin.

### Reaching the Origin
Once the particle crosses the peak, it accelerates towards the origin because the attractive force of the closer negative charges starts to win. To find its kinetic energy at the origin, we first need the potential there. Setting $x = 0$ in our potential equation gives:
$$V_0 \approx 2.4 	imes 10^4	ext{ V}$$

Notice that the potential at the origin is lower than the peak! The particle has fallen into a shallow "valley" between the peaks. Applying energy conservation once more, the total energy remains constant:
$$K_{	ext{origin}} + q_0 V_0 = q_0 V_{	ext{max}}$$

Solving for the kinetic energy at the origin:
$$K_{	ext{origin}} = q_0(V_{	ext{max}} - V_0) = 3 	imes 10^{-4}	ext{ J}$$

**This beautiful interplay of forces shows how a particle can navigate a complex potential landscape, slowing down as it climbs the barrier and speeding up as it falls into the well.**

Visualized Solution

$Setup$

$V (x)$

$Substitution$

$\frac{d V}{d x} = 0$

$x = \frac{5}{2}$

$V_{max}$

$v_{0}$

$V_{0}$

$K_{origin}$

$Conclusion$

The Sigma Insight: Electric Potential and Potential Difference

Analyzing the Setup Imagine a particle embarking on a journey along the $x$ -axis, starting from infinity and heading towards the origin

The Potential Landscape To understand this mountain, we need to map out the electric potential $V (x)$ along the $x$ -axis

Finding the Peak of the Mountain To find the minimum speed required to reach the origin, we must find the highest point of this potential mountain

The Minimum Speed to Cross Our particle starts at infinity, where the potential is zero

Reaching the Origin Once the particle crosses the peak, it accelerates towards the origin because the attractive force of the closer negative charges starts to win

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V(x)

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=25​​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V0​

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

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