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Chapter 23: Problem 62

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?

Short Answer

Expert verified
The potential difference required is approximately 4.79 V.

Step by step solution

01

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掳 angle with the vertical.
02

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掳 with the vertical, thus splitting the tension into vertical and horizontal components.
03

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, \(T = \frac{mg}{\cos(30^{\circ})}\).
04

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:\[E = \frac{mg\tan(30^{\circ})}{q}\]
05

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 \(\tan(30^{\circ}) = \frac{1}{\sqrt{3}}\).\[E = \frac{1.50 \times 10^{-3} \times 9.81 \times \frac{1}{\sqrt{3}}}{8.90 \times 10^{-6}}\approx 95.8\, \text{N/C}\]
06

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:\[V = 95.8 \times 0.05 \approx 4.79\, \text{V}\]
07

Conclusion

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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掳 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 蟽 (+蟽) and the other negative 蟽 (-蟽).
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掳 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.

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Most popular questions from this chapter

Two plastic spheres, each carrying charge uniformly distributed throughout its interior, are initially placed in contact and then released. One sphere is 60.0 \(\mathrm{cm}\) in diameter, has mass \(50.0 \mathrm{g},\) and contains \(-10.0 \mu \mathrm{C}\) of charge. The other sphere is 40.0 \(\mathrm{cm}\) in diameter, has mass \(150.0 \mathrm{g},\) and contains \(-30.0 \mu \mathrm{c}\) of charge. Find the maximum acceleration and the maximum speed achieved by each sphere (relative to the fixed point of their initial location in space), assuming that no other forces are acting on them. (Hint: The uniformly distributed charges behave as though they were concentrated at the centers of the two spheres.)

A point charge \(q_{1}=+5.00 \mu \mathrm{C}\) is held fixed in space. From a horizontal distance of \(6.00 \mathrm{cm},\) a small sphere with mass \(4.00 \times 10^{-3} \mathrm{kg}\) and charge \(q_{2}=+2.00 \mu \mathrm{C}\) is fired toward the fixed charge with an initial speed of 40.0 \(\mathrm{m} / \mathrm{s} .\) Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is 25.0 \(\mathrm{m} / \mathrm{s} ?\)

A metal sphere with radius \(R_{1}\) has a charge \(Q_{1}\) . Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conduct- ing wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

A charge of 28.0 \(\mathrm{nC}\) is placed in a uniform electric field that is directed vertically upward and has a magnitude of \(4.00 \times 10^{4} \mathrm{V} / \mathrm{m} .\) What work is done by the electric force when the charge moves (a) 0.450 \(\mathrm{m}\) to the right; (b) 0.670 \(\mathrm{m}\) upward; (c) 2.60 \(\mathrm{m}\) at an angle of \(45.0^{\circ}\) downward from the horizontal?

A total electric charge of 3.50 \(\mathrm{nC}\) is distributed uniformly over the surface of a metal sphere with a radius of 24.0 \(\mathrm{cm}\) . If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) \(48.0 \mathrm{cm} ;\) (b) \(24.0 \mathrm{cm} ;(\mathrm{c}) 12.0 \mathrm{cm}.\)
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