Skip to main content
Ch 25: The Electric Potential
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 25: The Electric PotentialProblem 48a
Chapter 25, Problem 48a

A room with 3.0-m-high ceilings has a metal plate on the floor with V=0 V and a separate metal plate on the ceiling. A 1.0 g glass ball charged to +4.9 nC is shot straight up at 5.0 m/s. How high does the ball go if the ceiling voltage is +3.0×106 V?

Verified step by step guidance
1
Determine the electric field between the plates. The electric field (E) is uniform and can be calculated using the formula: E = \(\frac{\Delta V}{d}\), where \(\Delta\) V is the potential difference between the plates and d is the distance between them. Here, \(\Delta\) V = 3.0 \(\times\) 10^6 \ \(\text{V}\) and d = 3.0 \ \(\text{m}\).
Calculate the force acting on the charged ball due to the electric field. The force F is given by F = qE, where q is the charge on the ball and E is the electric field calculated in the previous step. Here, q = 4.9 \(\times\) 10^{-9} \ \(\text{C}\).
Determine the net acceleration of the ball. The net force acting on the ball is the sum of the gravitational force (F_g = mg) and the electric force (F_e = qE). Use Newton's second law, F_{\(\text{net}\)} = ma, to find the net acceleration a. Here, m = 1.0 \ \(\text{g}\) = 0.001 \ \(\text{kg}\).
Use kinematic equations to find the maximum height. The ball's initial velocity is v_0 = 5.0 \ \(\text{m/s}\), and the final velocity at the maximum height is v = 0 \ \(\text{m/s}\). Use the equation v^2 = v_0^2 + 2a\(\Delta\) y to solve for \(\Delta\) y, the vertical displacement of the ball.
Add the initial height of the ball (if any) to the calculated displacement \(\Delta\) y to find the total height the ball reaches. In this case, the ball starts from the floor, so the total height is simply \(\Delta\) y.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Potential and Voltage

Electric potential, or voltage, is the amount of electric potential energy per unit charge at a point in an electric field. In this scenario, the ceiling has a voltage of +3.0×10^6 V, which creates an electric field that influences the motion of the charged glass ball. Understanding how voltage affects charged objects is crucial for predicting the ball's behavior in the electric field.
Recommended video:
Guided course
07:33
Electric Potential

Kinematics of Projectile Motion

Kinematics involves the study of motion without considering the forces that cause it. The glass ball is projected upward with an initial velocity of 5.0 m/s, and its motion can be analyzed using kinematic equations. These equations help determine the maximum height reached by the ball, factoring in its initial velocity and the effects of gravity.
Recommended video:
Guided course
04:44
Introduction to Projectile Motion

Energy Conservation in Electric Fields

The principle of energy conservation states that energy cannot be created or destroyed, only transformed. In this context, the kinetic energy of the glass ball is converted into potential energy as it rises in the electric field created by the ceiling's voltage. Understanding how energy transforms between kinetic and potential forms is essential for calculating the maximum height the ball can reach.
Recommended video:
Guided course
03:16
Intro to Electric Fields
Related Practice
Textbook Question

A 0.25 pg dust particle with 50 excess electrons is sitting at rest on top of a 5.0-cm-diameter metal sphere. Closing a switch charges the sphere almost instantaneously. To what potential must the sphere be charged to launch the dust particle to a height of 5.0 m? Ignore air resistance.

1
views
Textbook Question

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.2 cm apart with a 25 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential. With what speed does an electron exit the electron gun if its entry speed is close to zero?

1
views
Textbook Question

The electric potential in a region of space is given by V=V₀[(x²+2y²)/(0.10 m)²], where V₀ is a constant. A proton released from rest at (x, y)=(20 cm, 0 cm) reaches the origin with a speed of 7.5×105 m/s. At what value of y on the y-axis should a He+ ion (charge +e, mass 4 u) be released from rest to reach the origin with the same speed?

1
views
Textbook Question

An arrangement of source charges produces the electric potential V=5000x2 along the x-axis, where V is in volts and x is in meters. What is the maximum speed of a 1.0 g, 10 nC charged particle that moves in this potential with turning points at ±8.0 cm?

1
views
Textbook Question

Living cells 'pump' singly ionized sodium ions, Na+, from the inside of the cell to the outside to maintain a membrane potential ΔVmembrane=Vin−Vout=−70 mV. It is called pumping because work must be done to move a positive ion from the negative inside of the cell to the positive outside, and it must go on continuously because sodium ions 'leak' back through the cell wall by diffusion. At rest, the human body uses energy at the rate of approximately 100 W to maintain basic metabolic functions. It has been estimated that 20% of this energy is used to operate the sodium pumps of the body. Estimate—to one significant figure—the number of sodium ions pumped per second.

1
views
Textbook Question

What is the escape speed of an electron launched from the surface of a 1.0-cm-diameter glass sphere that has been charged to 10 nC?