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. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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.

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: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. Coulomb’s Constant

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB E-field near a point- charge Q is just most of the force formula 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.

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.

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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.

Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

q2q2 q1q1 x=0 x=0.2mx=-0.3m x Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

q2q2 q1q1 x=0 x=0.2mx=-0.3m 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. x Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

q2q2 q1q1 x=0 x=0.2mx=-0.3m This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q 1 will produce an E-field vector that points left, and q 2 gives an E-field vector to the right. E1E1 E2E2 x Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

q2q2 q1q1 x=0 x=0.2mx=-0.3m This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q 1 will produce an E-field vector that points left, and q 2 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. x Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB E1E1 E2E2

q2q2 q1q1 x=0 x=0.2mx=-0.3m x Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-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. At the origin, q 1 will produce an E-field vector that points left, and q 2 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. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB E1E1 E2E2

q2q2 q1q1 x=0 x=0.2mx=-0.3m E total (This means 400 N/C in the negative x-direction) x Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-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. At the origin, q 1 will produce an E-field vector that points left, and q 2 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. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

q2q2 q3q3 q1q1 x=0 x=0.2mx=-0.3m For part b) all we need to do is multiply the E-field from part a) times the new charge q 3. x Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-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. At the origin, q 1 will produce an E-field vector that points left, and q 2 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. (This means 400 N/C in the negative x-direction) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB E total

q2q2 q3q3 q1q1 x=0 x=0.2mx=-0.3m Note that this force is to the right, which is opposite the E-field This is because q 3 is a negative charge: E-fields are always set up as if there are positive charges. x Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-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. At the origin, q 1 will produce an E-field vector that points left, and q 2 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. (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 q 3. F on3 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB E total

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

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

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 Canceling and cross-multiplying, we get 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 Canceling and cross-multiplying, we get We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance. 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

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 2 nd law says that F net = 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 2 nd law says that F net = 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 2 nd law says that F net = 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 2 nd law says that F net = 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 2 nd law says that F net = 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 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB -4nC x=0x=0.8m +6nC x

-4nC x=0x=0.8m +6nC x 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. 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 For part a) which direction do the E-field vectors point? 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 x=0x=0.8m +6nC x 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 x=0x=0.8m +6nC x Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

a For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E1E1 E2E2 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 x=0x=0.8m +6nC x Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC

a For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E1E1 E2E2 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 x=0x=0.8m +6nC x Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC

a For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E1E1 E2E2 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 x=0x=0.8m +6nC x Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC For part b) E 1 points left and E 2 points right b E2E2 E1E1 Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC

a For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E1E1 E2E2 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 x=0x=0.8m +6nC x Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC For part b) E 1 points left and E 2 points right b E2E2 E1E1 Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC

a For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E1E1 E2E2 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 x=0x=0.8m +6nC x Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC For part b) E 1 points left and E 2 points right For part c) E 1 points right and E 2 points left c E1E1 E2E2 Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC b E2E2 E1E1 Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC

a For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E1E1 E2E2 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 x=0x=0.8m +6nC x Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC For part b) E 1 points left and E 2 points right For part c) E 1 points right and E 2 points left c E1E1 E2E2 Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC b E2E2 E1E1 Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC

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

x y Part a): TRY DRAWING THE E-FIELD VECTORS ON THE DIAGRAM 12 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 12 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 a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. E total = 0 E1E1 E2E2

x y x 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) E1E1 E2E2 Part b): both vectors point away from their charge, to the right. Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. E total = 0 E1E1 E2E2

x y x y 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) E1E1 E2E2 Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. E total = 0 Positive x-direction E1E1 E2E2 Part b): both vectors point away from their charge, to the right.

x y Part c): both vectors point away from their charge. We will need to use vector components to add them together. 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) 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 12 E 1,y (0.15,- 0.4) (0.15,0)(- 0.15,0) 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 12 E 1,y (0.15,- 0.4) (0.15,0)(- 0.15,0) 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 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) E2E2 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. E 1,y

x y 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) The 0.5m in this formula for E 2 is the distance to charge 2, using Pythagorean theorem or from recognizing a right triangle when you see it. 0.4m 0.3m 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) E2E2 E 1,y Part c): both vectors point away from their charge. We will need to use vector components to add them together.

x y 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) The 0.5m in this formula for E 2 is the distance to charge 2, using Pythagorean theorem or from recognizing a right triangle when you see it. 0.4m 0.3m 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) E 1,y Part c): both vectors point away from their charge. We will need to use vector components to add them together. E 2,x E 2,y

x y 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) The 0.5m in this formula for E 2 is the distance to charge 2, using Pythagorean theorem or from recognizing a right triangle when you see it. 0.4m 0.3m 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. E 1,y E 2,x E 2,y

x y 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) The 0.5m in this formula for E 2 is the distance to charge 2, using Pythagorean theorem or from recognizing a right triangle when you see it. 0.4m 0.3m 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. Add together the x-components and the y-components separately: E 1,y E 2,x E 2,y

x y 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) The 0.5m in this formula for E 2 is the distance to charge 2, using Pythagorean theorem or from recognizing a right triangle when you see it. 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. Add together the x-components and the y-components separately: Now find the magnitude and the angle using right triangle rules: 75.7º E total

x y Part d): TRY THIS ONE ON YOUR OWN FIRST (0.15,0)(- 0.15,0) 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) (0,0.2)

x y Part d): both vectors point away from their charge. We will need to use vector components to add them together. 12 E1E1 E2E2 (0,0.2) (0.15,0)(- 0.15,0) 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. From symmetry, we can see that E 2 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 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 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 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. We can apply our formula directly here:

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Electric Flux Try this example: 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 ε 0, 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.