Chapter 4: Q44E (page 779)

In a certain region of space the electric potential is given by $V = {Ax}^{2} y - {Bxy}^{2}$ , where $A = 5.00 V / m^{3}$ and $B = 8.00 V / m^{3}$ . Calculate the magnitude and direction of the electric field at the point in the region that has coordinates x = 2.00 m, y = 0.400 m, and z = 0.

Short Answer

Expert verified

$E = - 6.72 i - 7.2 j N / C E = 9.85 N / C θ = 227^{\circ}$

Step by step solution

parameters.

$V = {Ax}^{2} y - {Bxy}^{2} A = 5.00 V / m^{3} B = 8.00 V / m^{3} x = 2.00 m y = 0.4 m$

Calculating the magnitude and direction of the electric field.

The electric field is given by

$E = - ∆ v = E_{x} $ + E_{y} $$

The electric field is equal to the negative gradient of the potential so

$E_{x} = - \frac{\partial V}{\partial x} = - 2 A x y + B y^{2}$

Also in y direction

$E_{y} = - \frac{\partial V}{\partial y} = - A x^{2} + 2 B x y$

The electric field is given by

$E = [- 2 A x y + B y^{2}] i^+ [- A x^{2} + 2 B x y] j^$

The electric field E(2.0,0.4,0) is given by

localid="1664280297072" $E (2.0, 0.4, 0) = [(- 2) (5) (2) (0.4) + (8) {(0.4)}^{2}] i $ + [- (5) {(2)}^{2} + 2 (8) (0.4) (2)] = - 6.72 i $ - 7.2 j $ N I C$

The magnitude E I given by

$E = \sqrt{(7.2)^{2} + (6.72)^{2})} = 9.85 N / C$

Since both components are negative the angle lies in the 3rd quadrant so add to magnitude of angle

$θ = 180 + \arctan \frac{13.6}{/ 6.72} = 227^{\circ}$

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.

In a certain region of space the electric potential is given by $V = {Ax}^{2} y - {Bxy}^{2}$ , where $A = 5.00 V / m^{3}$ and $B = 8.00 V / m^{3}$ . Calculate the magnitude and direction of the electric field at the point in the region that has coordinates x = 2.00 m, y = 0.400 m, and z = 0.

Short Answer

Step by step solution

parameters.

Calculating the magnitude and direction of the electric field.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Turning Points in Physics

Thermodynamics

Oscillations

Engineering Physics

Famous Physicists

Work Energy and Power

Study anywhere. Anytime. Across all devices.

In a certain region of space the electric potential is given byV=Ax2y-Bxy2, whereA=5.00V/m3andB=8.00V/m3. Calculate the magnitude and direction of the electric field at the point in the region that has coordinates x = 2.00 m, y = 0.400 m, and z = 0.

Short Answer

Step by step solution

parameters.

Calculating the magnitude and direction of the electric field.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Turning Points in Physics

Thermodynamics

Oscillations

Engineering Physics

Famous Physicists

Work Energy and Power

Study anywhere. Anytime. Across all devices.

In a certain region of space the electric potential is given by $V = {Ax}^{2} y - {Bxy}^{2}$ , where $A = 5.00 V / m^{3}$ and $B = 8.00 V / m^{3}$ . Calculate the magnitude and direction of the electric field at the point in the region that has coordinates x = 2.00 m, y = 0.400 m, and z = 0.