Problem 62 - University Physics with Modern Physics

Question

A small sphere with mass 1.50 g hangs by a thread between two parallel vertical plates 5.00 \(\mathrm{cm}\) apart (Fig. \(\mathrm{P} 23.62 ) .\) The plates are insulating and have uniform surface charge densities \(+\sigma\) and \(-\sigma .\) The charge on the sphere is \(q=8.90 \times 10^{-6} \mathrm{C} .\) What potential difference between the plates will cause the thread to assume an angle of \(30.0^{\circ}\) with the vertical?

Step-by-Step Solution

Answer

The potential difference required is approximately 4.79 V.

Step 1: Understand the Problem Setup

We have a charged sphere of mass 1.50 g hanging between two parallel plates with surface charge densities \(+\sigma\) and \(-\sigma\). The plates create a uniform electric field \(E\) across them. The sphere experiences an electric force \(F_e\) and a gravitational force \(F_g\). We need to find the potential difference between the plates that causes the thread to form a 30° angle with the vertical.

Step 2: Identify Forces Acting on the Sphere

The forces acting on the sphere are:1. Gravitational force \(F_g = mg\), where \(m = 1.50 \times 10^{-3}\, \text{kg}\) and \(g = 9.81\, \text{m/s}^2\).2. Electric force \(F_e = qE\), where \(q = 8.90 \times 10^{-6}\, \text{C}\).The tension in the thread provides the centripetal force that balances these two forces. The thread forms an angle of 30° with the vertical, thus splitting the tension into vertical and horizontal components.

Step 3: Set Up the Components of Forces

The tension \(T\) in the thread has two components:Vertical: \(T_v = T\cos(30^{\circ}) = mg\)Horizontal: \(T_h = T\sin(30^{\circ}) = qE\)From the vertical component, \(T = \frac{mg}{\cos(30^{\circ})}\).

Step 4: Solve for Electric Field E

From the horizontal component, equate the electric force to the horizontal tension component:\(qE = T\sin(30^{\circ}) = \frac{mg\sin(30^{\circ})}{\cos(30^{\circ})}\)This gives the electric field:\[E = \frac{mg\tan(30^{\circ})}{q}\]

Step 5: Calculate Electric Field E

Substitute the known values:\(m = 1.50 \times 10^{-3}\, \text{kg}\), \(g = 9.81\, \text{m/s}^2\), \(q = 8.90 \times 10^{-6}\, \text{C}\), and \(\tan(30^{\circ}) = \frac{1}{\sqrt{3}}\).\[E = \frac{1.50 \times 10^{-3} \times 9.81 \times \frac{1}{\sqrt{3}}}{8.90 \times 10^{-6}}\approx 95.8\, \text{N/C}\]

Step 6: Find the Potential Difference

The relationship between the electric field \(E\) and the potential difference \(V\) between the plates is given by:\[V = Ed\]where \(d = 0.05\, \text{m}\) is the distance between the plates.Substituting \(E\) into the equation gives:\[V = 95.8 \times 0.05 \approx 4.79\, \text{V}\]

Step 7: Conclusion

The potential difference required between the plates to make the thread assume an angle of 30.0° with the vertical is approximately 4.79 V.

Key Concepts

Potential Difference

The concept of potential difference is crucial in understanding electric fields and their effects. It is the work needed to move a unit charge from one point to another in an electric field. In our setup, the small sphere experiences an electric force due to the potential difference between the charged plates.
This potential difference is responsible for creating an electric field that exerts force on the charged sphere. When the thread is at a 30° angle, this indicates that the force from the electric field balances with gravitational pull. To find this balance numerically, we use the relationship:

• Potential difference (V:

\[ V = Ed \]where * E is the electric field intensity in newtons per coulomb (N/C) * d is the distance between the plates in meters. Substitute the electric field and distance values to find the voltage required to maintain the sphere's position at the desired angle.

Surface Charge Density

Surface charge density describes how much charge is distributed over a unit area of the plates' surface. This concept is fundamental in generating the electric field that influences our hanging sphere. Surfaces with different charge densities produce varying electric field strengths, which in turn affect the potential difference. In our example:

• The plates have uniform surface charge densities: one with a positive σ (+σ) and the other negative σ (-σ).

These densities enable the plates to maintain a constant electric field across them. By understanding that surface charge density affects the magnitude of the electric field (E), it becomes clearer why changing this density alters the potential difference needed for the sphere's desired position. The uniformity assures a consistent electric force acting on the sphere.

Force Components

The scenario involves dissecting forces acting on the sphere into manageable components. The forces include both the gravitational force pulling downward and the electric force pushing perpendicular to the plates.

• The gravitational force \( F_g = mg \) acts vertically downward.
• The electric force \( F_e = qE \) acts horizontally due to opposing charges on the plates.

As the thread forms a 30° angle, it's important to split the thread's tension into two parts:* Vertical component (\( T_v = T\, \cos(30^\circ) \): keeps the sphere suspended against gravity.* Horizontal component (\( T_h = T\, \sin(30^\circ) \): balances the electric force.Understanding these force components helps students grasp how they interplay to stabilize the sphere in equilibrium as described in the exercise.