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Coulomb's Force COULOMB’S FORCE LAW Two point charges Multiple point charges Attractive Repulsive - - + + + The force exerted by one point charge on another acts along the line joining the charges. It varies inversely as the square of the distance separating the charges and is proportional to the product of the charges. The force is repulsive if the charges have the same sign and attractive if the charges have opposite signs. q1 Two point charges q1 and q2 q2 [F]-force; Newtons {N} [q]-charge; Coulomb {C} origin [r]-distance; meters {m} [ε]-permittivity; Farad/meter {F/m} Property of the medium COULOMB FORCE UNIT VECTOR Permittivity is a property of the medium. Also known as the dielectric constant. Permittivity of free space Coulomb’s constant Permittivity of a medium Relative permittivity For air FORCE IN MEDIUM SMALLER THAN FORCE IN VACUUM Insert oil drop Viewing microscope Eye Metal plates Millikan oil drop experiment Charging by contact Example (Question): A negative point charge of 1µC is situated in air at the origin of a rectangular coordinate system. A second negative point charge of 100µC is situated on the positive x axis at the distance of 500 mm from the origin. What is the force on the second charge? Example (Solution): A negative point charge of 1µC is situated in air at the origin of a rectangular coordinate system. A second negative point charge of 100µC is situated on the positive x axis at the distance of 500 mm from the origin. What is the force on the second charge? Y q1 = -1 µC origin X q2= -100 µC Example (Solution): Y q1 = -1 µC origin X q2= -100 µC END Multiple point charges It has been confirmed experimentally that when several charges are present, each exerts a force given by on every other charge. The interaction between any two charges is independent of the presence of all other charges. 5 point charges: Net force on q5 q1 q4 q5 q2 q3 Example (Question) The charging of individual raindrops is ultimately responsible for the electrical activity in thunderstorms. Suppose two drops with equal charge q are located on the x axis at ± a. Find the electric force on a third drop with charge Q located at an arbitrary point on the y axis. Example (Solution) The charging of individual raindrops is ultimately responsible for the electrical activity in thunderstorms. Suppose two drops with equal charge q are located on the x axis at ± a. Find the electric force on a third drop with charge Q located at an arbitrary point on the y axis. Y Q y q q X a a Example (Solution) Y Q y q q X a a Charge Q is the same distance r from the two other charges, so the force from each has the same magnitude: The direction of the two forces are different. Y Example (Solution) The x components cancel, Q while the y components add. y q q X a a Example (Solution) Y From figure Then Q y q q X a a Example (Solution) Y Consider the case where the charge is also located on the x axis, y = 0. Q q Q q a a X y q q X a a Example (Solution) Y Consider the case where Y the charge is a very large distance on the y axis such that y >> a. Q Q y q q X a a (2q) END Electric field Single point charge Multiple point charges Charge distribution We define an electric field similar to that of a gravitational field. A charge produces an electric field such that when another “test” charge is placed in the field it will experience an electrical force. - Electric field Test charge Electric force The electric field at any point is the force per unit charge experienced by a charge at that point. - Since the electric force is a vector then the electric field is also a vector. Electric field and electric force are vectors which point in the same direction when the test - charge q is positive. The electric field lines for a positive source charge point away from the source charge. - The electric field lines for a negative source charge point towards the source charge. Consider two point charges (+Q and +q) again and the force that exists between them. Electric force on q Electric field at q Q P Observation point Q [k]-Coulomb constant; meter/Farad {m/F} [q]-charge; Coulomb {C} [r]-distance; meters {m} [E]-Electric field; Newton/Coulomb {N/C} [E]-Electric field; Volt/meter {V/m} ELECTRIC FIELD (1) Electric field is a vector quantity. Thus at all points where the electric field exists it has magnitude and direction. (2) The charge q must be small and positive such that it does not disturb the source charge Q. (3) For a positive source charge Q the electric field vector and the electric force on the test charge q are in the same direction. (4) For a positive source charge Q, the electric field lines are directed away from the charge. (5) For a point charge Q located at the origin the electric field vector is: Y Q + X Example (Question) What is the electric field 30 cm from a charge q = 4.0 nC? Example (Solution) What is the electric field 30 cm from a charge q = 4.0 nC? Z P q Y X Example (Alternate Solution) Obtain the force on a test charge Z q Y X Then obtain the electric field END Electric field produced by multiple point charges (n = 2) q1 Z P q2 Y q1 and q2 are the source point charges. X P is the field point The charges (q1 and q2) produce the electric field observed at the point P Consider each charge in turn, independently of all other charges present. q1 Z P q2 Charge q1 produces an electric field at point P. Y Charge q2 produces an electric field at point P X The total electric field at P is the vector sum of the electric field produced by each individual charge. Consider charge q1 only q1 Z P q2 Electric field produced by charge q1 at P Y X Distance separating q1 and P Unit vector along line joining q1 and P Consider charge q2 only q1 Z P q2 Electric field produced by charge q2 at P Y X Distance separating q2 and P Unit vector along line joining q2 and P The total electric field at P is the vector sum of the electric field produced by each individual charge. q1 Z P q2 Y X Electric field produced by multiple point charges (n > 1) q1 q3 qi Z P q2 q q 4 The total electric field at P is the n vector sum of the electric field Y produced by each individual charge. X q5 Given a group of charges we find the net electric field at any point in space by using the principle of superposition. This is a general principle that says a net effect is the sum of the individual effects. Here, the principle means that we first compute the electric field at the point in space due to each of the charges, in turn. We then find the net electric field by adding these electric fields vectorially, as usual. Example (Question) The figure shows two point charges each of +10 nC separated in air by 8.0 m. Compute the electric field at the points A, B, and C + + Example (Solution) The figure shows two point charges each of +10 nC separated in air by 8.0 m. Compute the electric field at the points A, B, and C Point A: Make a sketch of the layout and then draw in vectors for the fields E1 produced by q1 and E2 produced by q2. To do that imagine a positive test charge at A. The force on it due to the charge q1 acts along the center-to-center line, is repulsive, and so points to the right. That means the E1 at A is to the right along the axis. Similarly, the force due to q2 on our imaginary test charge is to the left as is E2. Next calculate E1 and E2 and add them vectorially. We are spared this effort since E1 = E2, the two cancel and the field at A is zero + + Example (Solution) The figure shows two point charges each of +10 nC separated in air by 8.0 m. Compute the electric field at the points A, B, and C Point B: At point B the fields act as drawn in the figure, and we must find their components. First we will calculate E1 and E2. + + X Example (Solution) Point B: Since the charges and distances happen to be the same, the magnitudes of the two contributing fields are equal: Now for the vector components + + X Example (Solution) Point B: Since the charges and distances happen to be the same, the magnitudes of the two contributing fields are equal: N E1 = 2.81 (cos(45)xˆ + sin(45)yˆ ) ( C) N E 2 = 2.81 (−cos(45)xˆ + sin(45)yˆ ) ( C) € € + + X Example (Solution) Point B: The horizontal field components are equal and act in opposite direction. They will cancel. Only the vertical field components contribute, and in the same direction. N E = E sin 45o + E sin 45o = 2(2.81 )(0.707) B 1 ( ) 2 ( ) C € Y The direction is straight up in the positive y-direction + + X Example (Solution) Point C: The point C is similarly located with respect to the charges as point B is.The field magnitude at C is the same as at B except the direction is straight down in the negative y-direction.at The direction is + + straight down in the negative y-direction Example (Solution) The figure shows two point charges each of +10 nC separated in air by 8.0 m.
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