Physics 6B Electric Field Examples

Physics 6B Electric Field Examples
Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Electric charge can be either positive or negative.
Matter is chiefly comprised of electrons (negative), protons (positive) and neutrons (electrically neutral). A neutral object will have equal numbers of protons and electrons. Most of the time it is the negatively-charged electrons that can move back and forth between objects, so a negatively charged object has excess electrons, and a positively charged object has too few electrons. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Electric charge can be either positive or negative.
Matter is chiefly comprised of electrons (negative), protons (positive) and neutrons (electrically neutral). A neutral object will have equal numbers of protons and electrons. Most of the time it is the negatively-charged electrons that can move back and forth between objects, so a negatively charged object has excess electrons, and a positively charged object has too few electrons. Elementary charge: Charge is quantized, which means that the charge on any object is always a multiple the charge on a proton (or electron). e = 1.6 x C This is the smallest possible charge. Units for charge are Coulombs. The Coulomb is a very large unit, so you can expect to see tiny values like nano-Coulombs. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Charges interact with each other via the Electric Force.
Rules for interaction are based on the sign of the charge as follows: * Like charges repel * Opposite charges attract The force is given by Coulomb’s Law: Coulomb’s Constant Notice that this is just the magnitude of the force, and the r is the center-to-center distance between the two charges. My advice is to not put +- signs into this formula. Instead, find the direction of the force based on the attract/repel rules above. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Electric Fields This is just another (very important) way of looking at electric forces. We find the electric field near a charge distribution, then we can simply multiply by any charge to find the force on that charge. E-field near a point-charge Q is just most of the force formula Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Electric Field Lines The charge on the right is twice the magnitude of the charge on the left (and opposite in sign), so there are twice as many field lines, and they point towards the charge rather than away from it.

Electric Field of a Dipole
+q -q Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Electric Field of a Dipole
Notice that the field lines point away from positive and toward negative charges. This will always be true. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. q2 q1 x x=-0.3m x=0 x=0.2m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. q2 q1 x x=-0.3m x=0 x=0.2m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: E1 E2 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right. q2 q1 x x=-0.3m x=0 x=0.2m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: E1 E2 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right. This is how we can put the +/- signs on the E-fields when we add them up. q2 q1 x x=-0.3m x=0 x=0.2m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: Etotal This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right. This is how we can put the +/- signs on the E-fields when we add them up. q2 q1 x x=-0.3m x=0 x=0.2m (This means 400 N/C in the negative x-direction) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: Etotal This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right. This is how we can put the +/- signs on the E-fields when we add them up. q2 q3 q1 x x=-0.3m x=0 x=0.2m (This means 400 N/C in the negative x-direction) For part b) all we need to do is multiply the E-field from part a) times the new charge q3. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: Etotal Fon3 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right. This is how we can put the +/- signs on the E-fields when we add them up. q2 q3 q1 x x=-0.3m x=0 x=0.2m (This means 400 N/C in the negative x-direction) For part b) all we need to do is multiply the E-field from part a) times the new charge q3. Note that this force is to the right, which is opposite the E-field This is because q3 is a negative charge: E-fields are always set up as if there are positive charges. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two unequal charges repel each other with a force F
Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

The formula for electric force between 2 charges is
Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F? The formula for electric force between 2 charges is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F? The formula for electric force between 2 charges is If both charges are doubled, we will have Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F? The formula for electric force between 2 charges is If both charges are doubled, we will have So the new force is 4 times as large. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Formula for electric force between 2 charges is
Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F? Formula for electric force between 2 charges is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F? Formula for electric force between 2 charges is We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F? Formula for electric force between 2 charges is We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance. Canceling and cross-multiplying, we get Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F? Formula for electric force between 2 charges is We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance. Canceling and cross-multiplying, we get Square-roots of both sides gives us the answer: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Recall that Newton's 2nd law says that Fnet = ma.
When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released? Recall that Newton's 2nd law says that Fnet = ma. So this is really a problem about the force on the heavier charge. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released? Recall that Newton's 2nd law says that Fnet = ma. So this is really a problem about the force on the heavier charge. formula for electric force between 2 charges is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released? Recall that Newton's 2nd law says that Fnet = ma. So this is really a problem about the force on the heavier charge. formula for electric force between 2 charges is If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released? Recall that Newton's 2nd law says that Fnet = ma. So this is really a problem about the force on the heavier charge. formula for electric force between 2 charges is If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong: We cancel common terms and cross-multiply to get Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released? Recall that Newton's 2nd law says that Fnet = ma. So this is really a problem about the force on the heavier charge. formula for electric force between 2 charges is If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong: We cancel common terms and cross-multiply to get Square-root of both sides: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm -4nC +6nC x x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC +6nC x x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC +6nC x x=0 x=0.8m For part a) which direction do the E-field vectors point? -4nC +6nC a x x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Call the -4nC charge #1 and the +6nC charge #2
A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC +6nC x x=0 x=0.8m For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E2 Q1 = -4nC E1 Q2 = +6nC a x x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC +6nC x x=0 x=0.8m For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E2 Q1 = -4nC E1 Q2 = +6nC a x x=0 x=0.8m E2 Q1 = -4nC Q2 = +6nC E1 For part b) E1 points left and E2 points right b x x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC +6nC x x=0 x=0.8m For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E2 Q1 = -4nC E1 Q2 = +6nC a x x=0 x=0.8m E2 Q1 = -4nC Q2 = +6nC E1 For part b) E1 points left and E2 points right b x x=0 x=0.8m E2 E1 Q1 = -4nC Q2 = +6nC c x For part b) E1 points right and E2 points left x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the magnitude and direction of the net electric field at:
A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) x y Part a): TRY DRAWING THE E-FIELD VECTORS ON THE DIAGRAM 2 1 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) x y E1 E2 Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. Etotal = 0 2 1 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) x y E1 E2 Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. Etotal = 0 2 1 x y Part b): both vectors point away from their charge. E1 2 1 E2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) x y E1 E2 Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. Etotal = 0 2 1 x y Part b): both vectors point away from their charge. Positive x-direction E1 2 1 E2 Positive x-direction Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. x y (- 0.15,0) (0.15,0) 2 1 (0.15,- 0.4) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. x y (- 0.15,0) (0.15,0) 2 1 (0.15,- 0.4) E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. x y (- 0.15,0) (0.15,0) 2 1 (0.15,- 0.4) E2 E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. x y (- 0.15,0) (0.15,0) 2 1 The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a right triangle when you see it. 0.4m (0.15,- 0.4) 0.3m E2 E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. x y (- 0.15,0) (0.15,0) 2 1 The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a right triangle when you see it. 0.4m (0.15,- 0.4) E2,x 0.3m E2,y E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. x y (- 0.15,0) (0.15,0) 2 1 The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a right triangle when you see it. 0.4m (0.15,- 0.4) E2,x 0.3m E2,y Add together the x-components and the y-components separately: E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. x y (- 0.15,0) (0.15,0) 2 1 The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a right triangle when you see it. (0.15,- 0.4) 75.7º Add together the x-components and the y-components separately: Etotal Now find the magnitude and the angle using right triangle rules: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part d): TRY THIS ONE ON YOUR OWN FIRST... x y (0,0.2) (- 0.15,0) (0.15,0) 2 1 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part d): both vectors point away from their charge. We will need to use vector components to add them together. x y E1 E2 The 0.25m in this formula is the distance to each charge using the Pythagorean theorem or from recognizing a right triangle when you see it. (0,0.2) (- 0.15,0) (0.15,0) 2 1 From symmetry, we can see that E2 will have the same components, except for +/- signs. Now we can add the components (the x-component should cancel out) The final answer should be N/C in the positive y-direction. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Electric Flux Field lines passing through a surface are called “flux”. To find the flux through a surface, multiply field strength times the area of the surface. Here is the formula: Notice that if the field is not perpendicular to the area we need to multiply by cos(Φ). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Electric Flux Try this example:
A uniform electric field points in the +x direction and has magnitude 2000 N/C. Find the electric flux through a circular window of radius 10cm, tilted at an angle of 30° to the x-axis, as shown below. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A uniform electric field points in the +x direction and has magnitude 2000 N/C. Find the electric flux through a circular window of radius 10cm, tilted at an angle of 30° to the x-axis, as shown below. We can apply our formula directly here: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A positive 3µC charge is surrounded by a spherical surface of radius 0.2m, as shown. Find the net electric flux through the sphere. Notice that the electric field lines intercept the sphere at right angles, so the angle in our flux formula is 0°. We can put in formulas for the electric field near a point charge, and the surface area for a sphere to arrive at a nice formula. It turns out that the flux only depends on the enclosed charge, and furthermore, if we write it in terms of permittivity ε0, we find the following: This is called Gauss’ Law, and it works out that we don’t even need a sphere – any closed surface will do. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Physics 6B Electric Field Examples

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Physics 6B Electric Field Examples

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Presentation on theme: "Physics 6B Electric Field Examples"— Presentation transcript:

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