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UNIT 9 Electrostatics and Currents

UNIT 9 Electrostatics and Currents. Thursday March 15 th. Electrostatics and Currents. Thursday , March 15. TODAY’S AGENDA. Electric Field Hw : Practice D ( all ) p575. UPCOMING…. Thurs : Electric Field Problem Fri: NO SCHOOL (Teacher In-Service)

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UNIT 9 Electrostatics and Currents

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  1. UNIT 9Electrostatics and Currents

  2. Thursday March 15th Electrostatics and Currents

  3. Thursday, March 15 TODAY’S AGENDA • Electric Field • Hw: Practice D (all) p575 UPCOMING… • Thurs: Electric Field Problem • Fri: NO SCHOOL (Teacher In-Service) • Mon: Electric Potential Energy

  4. Chapter 16-17 Electrostatics and Current

  5. Section 3 The Electric Field Chapter 16 Electric Field Strength • Anelectric fieldis a region where an electric force on a test charge can be detected. • The SI units of the electric field, E, are newtons per coulomb (N/C). • The direction of the electric field vector, E,is in the direction of the electric force that would be exerted on a small positive test charge.

  6. Section 3 The Electric Field Chapter 16 Electric Field Strength, continued • Electric field strength depends on charge and distance. An electric field exists in the region around a charged object. • Electric Field Strength Due to a Point Charge

  7. Section 3 The Electric Field Chapter 16 Sample Problem Electric Field Strength A charge q1 = +7.00 µC is at the origin, and a charge q2 = –5.00 µC is on the x-axis 0.300 m from the origin, as shown at right. Find the electric field strength at point P,which is on the y-axis 0.400 m from the origin.

  8. Section 3 The Electric Field Chapter 16 Sample Problem, continued Electric Field Strength 1.Define the problem,andidentifytheknownvariables. Given: q1= +7.00 µC = 7.00  10–6 C r1 = 0.400 m q2 = –5.00 µC = –5.00 10–6 C r2 = 0.500 m • = 53.1º Unknown: E at P (y = 0.400 m) Tip:Apply the principle of superposition. You must first calculate the electric field produced by each charge individually at point P and then add these fields together as vectors.

  9. Section 3 The Electric Field Chapter 16 Sample Problem, continued Electric Field Strength 2. Calculate the electric field strength produced by each charge. Because we are finding the magnitude of the electric field, we can neglect the sign of each charge.

  10. Section 3 The Electric Field Chapter 16 Sample Problem, continued Electric Field Strength 3. Analyze the signs of the charges. The field vector E1 at P due to q1 is directed vertically upward, as shown in the figure, because q1 is positive. Likewise, the field vector E2at P due to q2 is directed toward q2 because q2 is negative.

  11. Section 3 The Electric Field Chapter 16 Sample Problem, continued Electric Field Strength 4. Find the x and y components of each electric field vector. For E1:Ex,1 = 0 N/C Ey,1 = 3.93  105 N/C For E2:Ex,2= (1.80  105 N/C)(cos 53.1º) = 1.08 105 N/C Ey,1= (1.80  105 N/C)(sin 53.1º)= –1.44 105 N/C

  12. Section 3 The Electric Field Chapter 16 Sample Problem, continued Electric Field Strength 5. Calculate the total electric field strength in both directions. Ex,tot = Ex,1 + Ex,2= 0 N/C + 1.08  105 N/C = 1.08  105 N/C Ey,tot = Ey,1 + Ey,2= 3.93  105 N/C – 1.44  105 N/C = 2.49  105 N/C

  13. Section 3 The Electric Field Chapter 16 Sample Problem, continued Electric Field Strength 6. Use the Pythagorean theorem to find the magnitude of the resultant electric field strength vector.

  14. Section 3 The Electric Field Chapter 16 Sample Problem, continued Electric Field Strength 7. Use a suitable trigonometric function to find the direction of the resultant electric field strength vector. In this case, you can use the inverse tangent function:

  15. Section 3 The Electric Field Chapter 16 Sample Problem, continued Electric Field Strength 8. Evaluate your answer. The electric field at point P is pointing away from the charge q1, as expected, because q1 is a positive charge and is larger than the negative charge q2.

  16. Section 3 The Electric Field Chapter 16 Electric Field Lines • The number of electric field lines is proportional to the electric field strength. • Electric field lines are tangent to the electric field vector at any point.

  17. Section 3 The Electric Field Chapter 16 Conductors in Electrostatic Equilibrium • The electric field is zero everywhere inside the conductor. • Any excess charge on an isolated conductor resides entirely on the conductor’s outer surface. • The electric field just outside a charged conductor is perpendicular to the conductor’s surface. • On an irregularly shaped conductor, charge tends to accumulate where the radius of curvature of the surface is smallest, that is, at sharp points.

  18. END

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