Chapter 19: Q14PE (page 696)

What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of $1.50 \times 10^{4} V$ ?

Short Answer

Expert verified

$1.50 \times 10^{6} V / m$ is the required electric field strength.

Step by step solution

Principle

The magnitude of the electric field (E) at any point is given by the potential gradient at that point.

$E = | - \frac{dV}{dr} |$

Here dV is the potential gradient and dr is the displacement from source charge.

The given data

The distance between the plates is:

$\begin{array}{rcl} d & = & (1.00 cm) (\frac{1 m}{100 cm}) \\ = & 0.0100 m \end{array}$

The potential difference between the two plates is: $ΔV = 1.50 \times 10^{4} V .$

Calculation of electric field strength

Equation (1) is used to calculate the electric field strength E between the two plates:

$E = \frac{ΔV}{d}$

After entering the numbers for $∆ V$ and d,

$\begin{array}{rcl} E & = & \frac{1.50 \times 10^{4} V}{0.0100 m} \\ = & 1.50 \times 10^{6} V / m \end{array}$

Therefore, the electric field strength is $1.50 \times 10^{6} V / m$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of $1.50 \times 10^{4} V$ ?

Short Answer

Step by step solution

Principle

The given data

Calculation of electric field strength

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Classical Mechanics

Kinematics Physics

Famous Physicists

Waves Physics

Modern Physics

Nuclear Physics

Study anywhere. Anytime. Across all devices.

What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×104V?

Short Answer

Step by step solution

Principle

The given data

Calculation of electric field strength

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Classical Mechanics

Kinematics Physics

Famous Physicists

Waves Physics

Modern Physics

Nuclear Physics

Study anywhere. Anytime. Across all devices.

What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of $1.50 \times 10^{4} V$ ?