Question

The maximum voltage at the center of a typical tandem electrostatic accelerator is 6.0 MV. If the distance from one end of the acceleration tube to the midpoint is 12 m, what is the magnitude of the average electric field in the tube under these conditions? (a) 41,000 V/m; (b) 250,000 V/m; (c) 500,000 V/m; (d) 6,000,000 V/m.

Short answer

The magnitude of the average electric field is 500,000 V/m, which is option (c).

Step-by-step solution

Understanding the Problem

We need to find the magnitude of the average electric field in an acceleration tube of a tandem electrostatic accelerator with a maximum voltage of 6.0 MV and a midpoint distance of 12 m.

Formula for the Electric Field

The electric field (E) is related to the voltage (V) and the distance (d) by the formula: \[ E = \frac{V}{d} \]This formula means that the average electric field is the voltage divided by the distance over which the voltage is applied.

Substitute Given Values

We are given a voltage (V) of 6.0 MV (which is 6,000,000 V) and a distance (d) of 12 m. Substitute these values into the formula: \[ E = \frac{6,000,000 \text{ V}}{12 \text{ m}} \]

Calculate the Electric Field

Calculate the electric field by dividing the voltage by the distance: \[ E = \frac{6,000,000}{12} = 500,000 \text{ V/m} \]

Conclusion and Answer Selection

The calculated electric field is 500,000 V/m, which matches option (c) from the given choices.

Key concepts

Understanding the Tandem Electrostatic Accelerator

A tandem electrostatic accelerator is a powerful device used in physics to accelerate charged particles to high speeds. It operates using high voltages to create strong electric fields that propel these particles through a vacuum tube. By using a series of electrodes, each with incrementally higher positive or negative charges, the particles gain energy as they move from one end of the accelerator to the other. The main goal is to increase the kinetic energy of the particles for use in scientific experiments, such as nuclear physics research or material analysis.
This type of accelerator can achieve very high voltages, often in the range of megavolts (MV), to provide the necessary force for this acceleration process. Located at the center of the accelerator, there is often a high-voltage terminal where maximum voltage is applied, allowing for efficient particle acceleration.

Grasping Voltage in Physics

Voltage, often referred to as electric potential difference, is a critical concept in electricity and electromagnetism. It is measured in volts (V) and represents the potential energy per unit charge that drives electric current through a circuit.
In the context of the tandem electrostatic accelerator, the voltage is crucial because it determines the strength of the electric field inside the acceleration tube. A higher voltage leads to a stronger electric field, which in turn provides a greater force to accelerate charged particles.

• One megavolt (MV) equals one million volts (1,000,000 V).
• In devices like accelerators, high voltages are necessary to impart significant energy to particles.

Thus, understanding voltage helps us appreciate how energy is transformed and utilized within these powerful scientific instruments.

Role of Distance in Electric Fields

Distance plays a vital role when calculating electric fields. The electric field (\( E \)) is influenced by the voltage applied across a given distance.
The relationship between electric field, voltage, and distance is defined by the equation: \[ E = \frac{V}{d} \] where \( V \) is the voltage and \( d \) is the distance over which the voltage is applied.
From this equation, we see that for a fixed voltage, a shorter distance results in a stronger electric field, and vice versa.

• In our exercise, the distance is the length from the midpoint to the end of the acceleration tube.
• Changing this distance alters the magnitude of the electric field and hence the force exerted on charged particles.

Clearly, distance is a fundamental factor affecting how electric fields exert force.

Effective Physics Problem Solving

Solving physics problems requires a structured approach to ensure accuracy and understanding. The exercise provided is an excellent example of this approach, dealing with concepts like electric fields in an electrostatic accelerator.
Here are essential steps to effectively tackle such problems:

• Understand the Problem: Clearly define what is being asked, identifying all given information and the required result.
• Identify Relevant Formulae: Use known relationships and equations that connect the given data to the desired result. In this case, \( E = \frac{V}{d} \) was used to find the electric field.
• Substitute and Solve: Insert the given values into the formula, performing the necessary calculations.
• Verify Your Results: Ensure the solution makes sense by checking the dimensions and the plausibility of the outcome.

By applying these principles, students can solve problems methodically and gain a deeper understanding of physics concepts.