A charged particle (mass m and charge q) moves along X-axis with velocity v_0. When it passes throug

Q: Why doesn't the horizontal velocity change inside the electric field?

The electric field is directed purely along the negative y-axis. Since force is a vector, it only affects motion in its own direction. There is no force component along the x-axis, so the horizontal velocity remains constant at $v_0$.

Q: Why is the path outside the field a straight line?

Once the particle crosses $x = d$, it exits the electric field. With no electric field, there is no force acting on the particle. According to Newton's First Law, an object in motion stays in motion with the same speed and in the same direction unless acted upon by a net force. Hence, it travels in a straight line.

Q: How do we know the slope of the straight line is $v_y / v_x$?

The slope of any path in the xy-plane is given by $dy/dx$. By the chain rule, $dy/dx = (dy/dt) / (dx/dt)$, which is exactly the vertical velocity $v_y$ divided by the horizontal velocity $v_x$.

Q: Can we solve this using the trajectory equation of a projectile?

Absolutely! The motion inside the field is mathematically identical to a horizontal projectile. The trajectory equation is $y = \frac{1}{2} a_y (x/v_x)^2$. You can find the position and slope at $x=d$ directly from this equation by substituting $x=d$ and taking the derivative $dy/dx$.

Q: What is the significance of the tangent intersecting at $d/2$?

This is a fundamental geometric property of parabolas with their vertex at the origin. If you draw a tangent at any point $(x_0, y_0)$ on the parabola $y = kx^2$, it will always intersect the x-axis at exactly $x_0 / 2$. It's a great shortcut to verify your straight-line equation!

Question

A charged particle (mass $m$ and charge $q$) moves along X-axis with velocity $v_0$. When it passes through the origin it enters a region having uniform electric field $\mathbf{E} = -E\hat{\mathbf{j}}$ which extends upto $x = d$. Equation of path of electron in the region ($x > d$) is

Options:

(A) $y = \frac{qEd}{mv_0^2}x$

(B) $y = \frac{qEd}{mv_0^2}(x-d)$

(C) $y = \frac{qEd}{mv_0^2}\left(\frac{d}{2} - x\right)$

(D) $y = \frac{qEd^2}{mv_0^2}x$

Sigma Percentile · Accepted Answer

The problem of a charged particle moving through a localized electric field is a classic in physics, beautifully bridging the concepts of electrostatics and kinematics. It is essentially the electrostatic equivalent of a ball thrown horizontally off a cliff!

### Analyzing the Setup
Imagine a charged particle, with mass $m$ and charge $q$, cruising along the x-axis with a constant velocity $v_0$. Everything is peaceful until it crosses the origin. Suddenly, it enters a region of width $d$ where a uniform electric field $\mathbf{E} = -E\hat{\mathbf{j}}$ is pointing straight down.

Just like gravity pulls a projectile downwards, this electric field exerts a constant downward force on our particle. Because there is no force in the horizontal direction, the horizontal velocity $v_x$ remains perfectly constant at $v_0$.

### The Parabolic Journey
Inside the electric field ($0 < x \le d$), the particle experiences a vertical acceleration:
$$a_y = \frac{-qE}{m}$$

Since the horizontal velocity is constant, the time it takes to cross this region is simply the distance divided by the speed:
$$t = \frac{d}{v_0}$$

By the time the particle reaches the exit boundary at $x = d$, it has acquired a vertical velocity. Using the first equation of motion ($v = u + at$), we find:
$$v_y = a_y t = \left(\frac{-qE}{m}\right)\left(\frac{d}{v_0}\right) = \frac{-qEd}{mv_0}$$

We also need to know exactly **where** it exits. Using the second equation of motion ($s = ut + \frac{1}{2}at^2$), the vertical displacement is:
$$y_1 = \frac{1}{2}a_y t^2 = \frac{1}{2}\left(\frac{-qE}{m}\right)\left(\frac{d}{v_0}\right)^2 = \frac{-qEd^2}{2mv_0^2}$$

### Breaking Free: The Straight Line
Once the particle crosses $x = d$, the electric field vanishes. With no forces acting on it, Newton's First Law dictates that the particle must continue in a straight line at a constant velocity.

The slope $m'$ of this straight line is simply the ratio of its velocity components:
$$m' = \frac{v_y}{v_x} = \frac{\frac{-qEd}{mv_0}}{v_0} = \frac{-qEd}{mv_0^2}$$

Now, we have a point $(d, y_1)$ and a slope $m'$. We can use the point-slope form of a straight line equation:
$$y - y_1 = m'(x - x_1)$$

Substituting our values:
$$y - \left(\frac{-qEd^2}{2mv_0^2}\right) = \frac{-qEd}{mv_0^2}(x - d)$$

### Final Calculation
Let's expand and simplify this equation. 
$$y + \frac{qEd^2}{2mv_0^2} = \frac{-qEd}{mv_0^2}x + \frac{qEd^2}{mv_0^2}$$

Moving the constant term to the right side:
$$y = \frac{-qEd}{mv_0^2}x + \frac{qEd^2}{mv_0^2} - \frac{qEd^2}{2mv_0^2}$$

Notice how $1 - \frac{1}{2} = \frac{1}{2}$. The $d^2$ terms combine beautifully:
$$y = \frac{-qEd}{mv_0^2}x + \frac{qEd^2}{2mv_0^2}$$

Finally, factoring out the common terms, we arrive at our elegant final equation:
$$y = \frac{qEd}{mv_0^2}\left(\frac{d}{2} - x\right)$$

**A Brilliant Geometric Insight:** If you look closely at the final equation, what happens when $y = 0$? You get $x = d/2$. This means if you extend the straight-line path backwards, it intersects the x-axis at exactly half the horizontal distance. This is a universal geometric property of all such parabolic trajectories!

Select Answer:

Visualized Solution

Analyzing the Setup

Trajectory Inside and Outside

Time Inside the Field

Velocity at Exit

Position at Exit

Slope of the Straight Line

Equation of the Straight Line

Final Simplification

The Geometric Insight

The Sigma Insight: Electric Field

Analyzing the Setup Imagine a charged particle, with mass $m$ and charge $q$ , cruising along the x-axis with a constant velocity $v_{0}$

The Parabolic Journey

Breaking Free

Final Calculation

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