Question
A 40.0-cm-diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be $5.20 \times 10^{5} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}$. What is the magnitude of the electric field?
Step 1
Here, $\Phi_E$ is the electric flux, $E$ is the electric field, $A$ is the area, and $\theta$ is the angle between the electric field and the normal to the area. For maximum electric flux, $\cos \theta = 1$ (as $\theta = 0$ degrees). Show more…
Show all steps
Your feedback will help us improve your experience
Aja S and 53 other Physics 102 Electricity and Magnetism educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A 40.0 -cm-diameter circular loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be $5.20 \times$ $10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} .$ What is the magnitude of the electric field?
A 40.0 -cm-diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is $5.20 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$ . What is the magnitude of the electric field?
A 40.0 -cm-diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be $5.20 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$ What is the magnitude of the electric field?
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD