The bob of a simple pendulum has mass 2 g and a charge of 5.0 uC. It is at rest in a uniform horizon

Q: Why do we resolve the tension instead of the other forces?

Resolving the tension aligns our coordinate system with the horizontal electric force and vertical gravitational force. This makes the math much simpler because we only have to break down one vector instead of two.

Q: Do I always need to convert grams to kilograms?

Yes! In physics, always convert mass to kilograms (the standard SI unit) to ensure your force calculations (like $mg$) result correctly in Newtons.

Q: What happens if the charge on the bob is negative?

If the charge were negative, the electric force would point in the opposite direction of the electric field. The pendulum would simply swing to the left instead of the right, but the magnitude of the angle would remain exactly the same.

Q: Why is the net force zero?

The problem states the pendulum is 'at rest in equilibrium,' which means it is not accelerating. According to Newton's second law ($F = ma$), zero acceleration means zero net force.

Q: Can I use $\sin\theta$ or $\cos\theta$ to find the angle instead of $\tan\theta$?

You could, but you would need to calculate the exact value of the tension $T$ first using the Pythagorean theorem. Using $\tan\theta$ allows us to divide the equations and completely eliminate $T$, saving us a lot of calculation time!

Question

The bob of a simple pendulum has mass $2	ext{ g}$ and a charge of $5.0\mu	ext{C}$. It is at rest in a uniform horizontal electric field of intensity $2000	ext{ V/m}$. At equilibrium, the angle that the pendulum makes with the vertical is (take $g = 10	ext{ m/s}^2$)

Options:

(A) $	an^{-1}(2.0)$

(B) $	an^{-1}(0.2)$

(C) $	an^{-1}(5.0)$

(D) $	an^{-1}(0.5)$

Sigma Percentile · Accepted Answer

### The Leaning Pendulum: Balancing Gravity and Electricity

Imagine a simple pendulum. Normally, it hangs straight down, right? Gravity pulls it directly towards the center of the Earth. But what if we introduce an invisible force—a uniform horizontal electric field—pulling it sideways? The pendulum will swing out and eventually come to rest at a new equilibrium angle. Let's dive into the physics of how to find that exact angle.

### Analyzing the Setup

Let's visualize the situation. The pendulum bob is displaced from the vertical by an angle $	heta$ because of the horizontal electric field. There are exactly three forces acting on it:

1. **Gravity:** The weight of the bob, $mg$, acting vertically downwards.
2. **Electric Force:** The force exerted by the electric field, $qE$, pulling it horizontally.
3. **Tension:** The force $T$ in the string pulling it back towards the pivot point.

Because the bob is at rest in equilibrium, the net force acting on it must be zero. This means the forces must perfectly balance each other out in every direction.

### The Master Equations

To make the math simple, we resolve the diagonal tension $T$ into vertical and horizontal components.

The vertical component of the tension, $T\cos	heta$, must balance the downward pull of gravity:
$T\cos	heta = mg$

The horizontal component of the tension, $T\sin	heta$, must balance the sideways pull of the electric field:
$T\sin	heta = qE$

Now, we have a beautiful system of two equations. To find the angle $	heta$, we can elegantly eliminate the tension $T$ by dividing the second equation by the first:

$\frac{T\sin	heta}{T\cos	heta} = \frac{qE}{mg}$

The $T$ cancels out, leaving us with a simple, powerful relationship:

$	an	heta = \frac{qE}{mg}$

### Final Calculation

Now, it's time to substitute the given values into our equation. **Crucially, we must convert all units to standard SI units** to avoid silly mistakes.

- The charge $q = 5.0 \mu	ext{C} = 5.0 	imes 10^{-6} 	ext{ C}$
- The electric field $E = 2000 	ext{ V/m}$
- The mass $m = 2 	ext{ g} = 2 	imes 10^{-3} 	ext{ kg}$
- The acceleration due to gravity $g = 10 	ext{ m/s}^2$

Plugging these into our master equation:

$	an	heta = \frac{5.0 	imes 10^{-6} 	imes 2000}{2 	imes 10^{-3} 	imes 10}$

Let's simplify the numerator and denominator:

$	an	heta = \frac{10 	imes 10^{-3}}{20 	imes 10^{-3}}$

$	an	heta = \frac{10^{-2}}{2 	imes 10^{-2}} = 0.5$

Finally, taking the inverse tangent gives us the equilibrium angle:

**$	heta = 	an^{-1}(0.5)$**

And there we have it! By carefully breaking down the forces and respecting our units, we've successfully found the resting angle of our electrified pendulum.

Animated Solution for Physics - Electrostatics: The bob of a simple pendulum has mass $2 g$ and a charge of $5.0 μ C$ . It is at rest in a uniform horizontal electric field of intensity $2000 V/m$ . At equilibrium, the angle that the pendulum makes with the vertical is (take $g = 10 m/s^{2}$ )

Select Answer:

Visualized Solution

$Free Body Diagram$

$Equilibrium Conditions$

$Finding tan θ$

$Substituting Values$

$Calculation$

$What’s Next?$

The Sigma Insight: Electric Field

The Leaning Pendulum

Analyzing the Setup

The Master Equations

Final Calculation

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