A uniform electric field, E = -400√3 y NC^-1 is applied in a region. A charged particle of mass m ca

Q: Why can we use standard projectile formulas here?

Because the electric field is uniform and acts only in the y-direction, it creates a constant downward acceleration just like gravity. The kinematics are mathematically identical.

Q: Why are there two different angles that hit the same target?

For any given range less than the maximum range, there are always two complementary angles of projection ($\theta$ and $90^\circ - \theta$) that will reach it. One is a lower, faster path, and the other is a higher, slower path.

Q: How do I handle the massive powers of 10 in the calculation?

Don't evaluate them immediately. Keep them as powers of 10 and substitute them into the final formula. You'll notice that the $10^{12}$ from $u^2$ perfectly cancels the $10^{12}$ in the acceleration.

Q: Why did we convert the time to microseconds?

The calculated time is in the order of $10^{-6}$ seconds. The prefix 'micro' ($\mu$) stands for $10^{-6}$, which makes the final answer much cleaner and matches the options provided.

Q: What if the charge was negative?

If the charge were negative, the electric force would act opposite to the electric field (upwards). The particle would accelerate upwards, and it would never hit the target on the x-axis!

Question

A uniform electric field, $\vec{E} = -400\sqrt{3}\hat{y}	ext{ NC}^{-1}$ is applied in a region. A charged particle of mass $m$ carrying positive charge $q$ is projected in this region with an initial speed of $2\sqrt{10} 	imes 10^6	ext{ ms}^{-1}$. This particle is aimed to hit a target T, which is $5	ext{ m}$ away from its entry point into the field as shown schematically in the figure. Take $\frac{q}{m} = 10^{10}	ext{ Ckg}^{-1}$. Then-

Options:

(A) the particle will hit T if projected at an angle $45^\circ$ from the horizontal

(B) the particle will hit T if projected either at an angle $30^\circ$ or $60^\circ$ from the horizontal

(C) time taken by the particle to hit T could be $\sqrt{\frac{5}{6}}\,\mu	ext{s}$ as well as $\sqrt{\frac{5}{2}}\,\mu	ext{s}$

(D) time taken by the particle to hit T is $\sqrt{\frac{5}{3}}\,\mu	ext{s}$

Sigma Percentile · Accepted Answer

### The Electric Gravity

Imagine you are throwing a ball, but instead of gravity pulling it down, it's a powerful electric field. The field points downwards, so our positively charged particle experiences a constant downward acceleration.

We know from Newton's second law that acceleration is force over mass. Here, the force is charge times electric field.

$$a_y = \frac{qE}{m}$$

Substituting the given values, we get an enormous acceleration:

$$a_y = (10^{10}) 	imes (400\sqrt{3}) = 4\sqrt{3} 	imes 10^{12} 	ext{ m/s}^2$$

Don't get intimidated by these large numbers, they will cancel out beautifully.

### The Range Equation

Now look at the setup. The particle needs to hit target T, which is exactly $5	ext{ m}$ away on the horizontal axis. This is a classic projectile motion problem!

We can directly use our standard range formula, just replacing $g$ with our new effective acceleration, $a_y$.

$$R = \frac{u^2 \sin 2	heta}{a_y}$$

Let's substitute the values and get the answer. We plug in the initial velocity $u$, and our calculated acceleration.

$$5 = \frac{(2\sqrt{10} 	imes 10^6)^2 \sin 2	heta}{4\sqrt{3} 	imes 10^{12}}$$

Notice how the $10^{12}$ terms in the numerator and denominator are perfectly set up to cancel each other out. This is where the magic happens.

$$5 = \frac{40 	imes 10^{12} \sin 2	heta}{4\sqrt{3} 	imes 10^{12}}$$

### Two Paths, One Destination

After simplifying, we are left with a very neat trigonometric equation.

$$5 = \frac{10}{\sqrt{3}} \sin 2	heta$$

Rearranging this, we find that $\sin 2	heta$ is exactly equal to $\frac{\sqrt{3}}{2}$.

$$\sin 2	heta = \frac{\sqrt{3}}{2}$$

There is a catch here. Sine is positive in both the first and second quadrants. So, $2	heta$ can be $60^\circ$, or $120^\circ$.

$$2	heta = 60^\circ \implies 	heta = 30^\circ$$

$$2	heta = 120^\circ \implies 	heta = 60^\circ$$

This gives us two possible angles of projection: $30^\circ$ and $60^\circ$. Both paths will hit the target!

### The Time of Flight

Since we have two different paths, the particle will take different amounts of time to reach the target depending on the angle. Let's bring back our time of flight formula to calculate these two distinct times.

$$T = \frac{2u \sin	heta}{a_y}$$

For the $30^\circ$ angle, we substitute $\sin 30^\circ$, which is $\frac{1}{2}$.

$$T_1 = \frac{2(2\sqrt{10} 	imes 10^6) 	imes (1/2)}{4\sqrt{3} 	imes 10^{12}}$$

Carefully crunching the numbers, we get $\sqrt{\frac{10}{12}} 	imes 10^{-6}	ext{ s}$, which simplifies beautifully to $\sqrt{\frac{5}{6}}\,\mu	ext{s}$. That's our first possible time.

Now for the steeper $60^\circ$ path. We use $\sin 60^\circ$, which is $\frac{\sqrt{3}}{2}$.

$$T_2 = \frac{2(2\sqrt{10} 	imes 10^6) 	imes (\sqrt{3}/2)}{4\sqrt{3} 	imes 10^{12}}$$

The $\sqrt{3}$ terms cancel out, and we are left with $\sqrt{\frac{10}{4}} 	imes 10^{-6}	ext{ s}$, simplifying to $\sqrt{\frac{5}{2}}\,\mu	ext{s}$. A higher angle means it stays in the air longer!

**Conclusion:** The particle can hit the target if projected at either $30^\circ$ or $60^\circ$. The corresponding times of flight match exactly with the values we found. This makes options (B) and (C) the correct choices.

Select Answer:

Visualized Solution

$F_{e} = q E$

$a_{y} = \frac{q E}{m}$

$R = \frac{u ^{2} s i n 2 θ}{a _{y}}$

$5 = \frac{u ^{2} s i n 2 θ}{a _{y}}$

$sin 2 θ = \frac{3}{2}$

$θ = 3 0^{\circ}, 6 0^{\circ}$

$T = \frac{2 u s i n θ}{a _{y}}$

$T_{1} (θ = 3 0^{\circ})$

$T_{2} (θ = 6 0^{\circ})$

$T_{1}, T_{2}$

The Sigma Insight: Electric Field

The Electric Gravity

The Range Equation

Two Paths, One Destination

The Time of Flight

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Select Answer:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=ay​u2sin2θ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5=ay​u2sin2θ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2θ=23​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=30∘,60∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=ay​2usinθ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​(θ=30∘)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​(θ=60∘)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​,T2​