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**Electric Potential Energy & Electric Potential Unit 8**

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**Recall ‘Work’ from earlier**

Work done by a force is given by: W = F d cos(q) or +W: Force is in direction moved -W: Force is opposite direction moved W=0: Force is 90o to direction moved Conservative Forces DU = -Wfield Use book or brick for prop. Ask students if I am doing positive or negative work, what about gravity.

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**Electric Potential Energy**

ΔU = -Wfield General Points: 1) Potential Energy increases if the particle moves in the direction opposite to the field force. Work will have to be done by an external agent for this to occur 2) Potential Energy decreases if the particle moves in the same direction as the field force on it

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**Electrical Potential Energy**

Graphical look at EPE The potential energy is taken to be zero when the two charges are infinitely separated

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**Potential Energy of a System of Charges**

Start by putting first charge in position No work is necessary to do this Next bring second charge into place Now work is done by the electric field of the first charge. This work goes into the potential energy between these two charges. Now the third & fourth charge are put into place Work is done by the electric fields of the two previous charges. The total EPE is then given by (signs matter)

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Example A test charge is brought separately to the vicinity of a positive charge Q at pt B A q r Q Charge +q is brought to pt A, a distance r from Q B A (a) UA < UB (b) UA = UB (c) UA > UB I) Compare the potential energy of q to that of Q. (a) (b) (c) II) Suppose charge q has mass m and is released from rest from the above position (a distance r from Q). What is its velocity vf as it approaches r = ∞ ?

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**Work Done by Uniform Electric Field**

Force on charge is Work is done on the charge by field

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ELECTRIC POTENTIAL Consider a ball of mass, m, placed at a point in space (height, h, above Earth). It would possess a certain PE per unit mass due to it being in the gravitational field of Earth. If the ball was replaced by a bowling ball of mass, M, it too, would possess the SAME potential energy per unit mass.

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Similarly, 2 different sized charges at the same distance from the charged sphere (green) will have the same EPE/charge. The electric potential energy per unit charge for some location in an electrical field is called electric potential.

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Electric Potential Electric potential is defined as the potential energy per unit charge at a point in space

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**Comparison: Electric Potential Energy vs. Electric Potential**

Electric Potential Energy (U) - the energy of a charge at some location. Electric Potential (V) - tells what the EPE would be if a charge were located there

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**Voltage and Potential Energy**

Positive charge will naturally move towards lower electrical potential energies, lower voltage.

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**General Points for either positive or negative charges:**

Positive potential is taken to be higher by definition due to positive test charge.

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**Relation between Potential and Field**

The work done by the electric field force in moving a charge from point a to point b is given by

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**Electric Potential for a Point Charge**

The direction of the electric field from a point charge is always radial. We integrate from distance r (distance from the point charge) along a radial line to infinity:

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**What is the electric potential difference between A and B?**

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Rank (a), (b) and (c) according to the net electric potential V produced at point P by two protons. Greatest first. A: (b), (c), (a) B: all equal C: (c), (b), (a) D: (a) and (c) tie, then (b)

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**Question… The electric potential at point A is _______ at point B**

greater than equal to less than

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**Points A, B, and C lie in a uniform electric field.**

1) If a positive charge is moved from point A to point B, its electric potential energy a) Increases b) decreases c) doesn’t change 2) Compare the potential differences between points A and C and points B and C. a) VAC > VBC b) VAC = VBC c) VAC < VBC

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Two Charges In region II (between the two charges) the electric potential is 1) always positive 2) positive at some points, negative at others. 3) always negative I II III Q=+7.0mC Q=-3.5 mC

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**Potential for two charges**

Calculate electric potential at point A due to charges A 4 m 6 m How much work do you have to do to bring a 2 mC charge from far away to point A? Q=+7.0mC Q=-3.5 mC

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**Potential of a solid conducting sphere (radius R) with charge +Q**

Find V at the following locations: +Q i) At r > R R ii) at r = R

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R iii) at r < R

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**E vs V graph for conductor**

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**V for uniformly charged, nonconducting sphere (radius R)**

a) Find ΔV for r > R moving from ∞ to a distance r from center

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**b) Find ΔV moving from ∞ to a distance r where r < R.**

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Equipontentials 3/25/2017

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**Equipotential (EP) Surfaces & Their Relation to Electric Field**

An equipotential surface is a surface on which the electric potential is the same everywhere. The EP surfaces that surround the point charge +q are spherical. The electric force does no work as a charge moves on a path that lies on an EP surface, such as the path ABC. However, work is done by the electric force when a charge moves along the path AD.

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**Equipotential Surfaces & Their Relation to Electric Field**

Equipotential surfaces (in blue) of an electric dipole. The surfaces are drawn so that at every point they are perpendicular to the electric field lines (in red) of the dipole. The radially directed electric field of a point charge is perpendicular to the spherical equipotential surfaces that surround the charge. The electric field points in the direction of decreasing potential.

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Consider two conducting spheres with differing radii Ra and Rb sitting on insulating stands far apart. The sphere with radius Ra has an electric charge +Q. If we connect a thin, conducting line between the spheres, then disconnect it, what are the charges on the spheres?

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**2 identical spheres question**

+Q +Q/ higher V? Attach wire btw spheres…what happens? What is final charge of each?

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**Potential difference between charged plates**

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**Capacitance + - Units = Farads (F)**

Describes how much charge an arrangement of conductors can hold for a given voltage applied. - When the battery is connected to the pair of plates, charges will flow until the top plate’s potential is the same as the + side of the battery, and the bottom plate’s potential is the same as the – side of the battery. No potential difference. Q is the amount of charge on a plate and ΔV is the voltage applied to the plates Units = Farads (F)

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**Work to charge conductor**

Consider a spherical uncharged conductor, radius R. After a small amount of charge is placed on conductor, its potential becomes V = kQ/R (where V∞=0). To further charge conductor, work must be done to bring additional increments of charge, dQ, to place on surface. W = ΔV dQ…the amount of work increases as each dQ is added and sphere becomes more charged.

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EPE of conductor

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**Capacitance for Parallel Plates**

The E field is constant The geometry is simple, only the area and plate separation are important. To calculate capacitance, we first need to determine the E-field between the plates. We did this using Gauss’ Law: separation d Total charge q on inside of plate E and dA parallel area A V+ V-

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**Need to find potential difference. Since E=constant**

Total charge q on inside of plate E and dA parallel V- V+ area A

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How large is 1 Farad? If a parallel plate capacitor has plates that are separated by only 1mm, in order to achieve 1F, the area of each plate would be… A = 1.1x108m2 This corresponds to about 6 miles on each side of plate! Obviously, this is impractical to achieve large capacitance. Therefore, what do they do?

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Energy per unit Volume It is necessary at times to relate energy per unit volume to electric field of capacitor (parallel plate)

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Example Charge 8.0uF capacitor (C1) by connecting it to a 120V potential difference. Now remove the power supply. a) Find charge on capacitor b) Find energy stored in capacitor

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**c) Now connect C1 to another capacitor, C2= 4.0uF, initially uncharged.**

What will be the potential difference across each capacitor & charge on each after equilibrium is reached? Conservation of charge

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**Method for finding C for various geometries of plates**

1) We are trying to calculate C where C = Q / ΔV 2) In order to acquire ΔV, we must use 3) Therefore we must find expression for E first using Gauss’ Law. 4) Find E, then ΔV, and then C.

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**Find capacitance of concentric cylindrical conductors with radius a (inside) & radius b.**

Inside charge is +, outside is - Field is radially outward

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End view

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**Spherical Capacitor (2 concentric spheres, inner radius a & outer radius b as shown)**

+Q -Q b a

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**integrate inward instead of outward this time because of the negative we get with 1/r^2**

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**Combination of Capacitors & Equivalent Capacitance (CEQ)**

1) Capacitors in Parallel & CEQ

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**Consider 3 identical capacitors in parallel connected to battery of voltage, V. Find CEQ**

All top plates are at same potential and so are bottom plates, so…

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Splitting capacitor into 3 separate capacitors in parallel all with equal potential difference btwn them (same as battery)

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**Capacitors in Series Charge on each capacitor is the same, Q.**

If you place 2 capacitors in series, the charge remains the same, but the potential difference is less for each capacitor Charge on each capacitor is the same, Q.

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**Consider the individual voltages across each capacitor**

Since q is the same for each The sum of these voltages is the total voltage of the battery, V

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**Combo Example Find Ceq, Q1, Q2, Q3, V1, V2, V3, & Qtotal**

A 12 battery is connected to the combination of capacitors as shown. 8uF C1 C2 C3 12V 2uF 4uF Find Ceq, Q1, Q2, Q3, V1, V2, V3, & Qtotal

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DIELECTRICS E If the strength of the electric field between the plates of an air filled capacitor becomes too strong, then the air can no longer insulate the charges from sparking (discharging) between the plates. For air, this breakdown occurs when the electric field is greater than 3x106 V/m. (this is what occurs during a lightning strike)…V/m is equivalent to N/C. In order to keep this from happening, an insulator, or dielectric, is often inserted between the plates to reduce the strength of the electric field, which yields a larger capacitance.

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**Why does dielectric reduce E?**

Dielectric material is polar and molecules polarize as shown. The charge alignment creates an E-field within the material which OPPOSES the original E-field between the plates. Electrical forces create a torque to rotate and align molecules

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**The dielectric is measured in terms of a dimensionless constant, κ (greek kappa) ≥ 1. (see table)**

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Assuming a capacitor is charged with no power source present, dielectric reduces E which reduces V (according to V = Ed) while d remains constant. If V reduces, then C increases (according to C = Q / V)

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example A parallel-plate air capacitor is charged by placing a 90-V battery across it. The battery is then removed. An insulating, dielectric fluid is inserted between the plates. The voltage across the capacitor is now 28V. What is the dielectric constant of the fluid?

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Example 2 A parallel-plate air capacitor holds a charge of 30nC when a voltage of V is placed across its plates. If the battery is not removed and a dielectric fluid is inserted between the plates, the charge on the plates increases to 87nC. Find the dielectric value.

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The charged plates of an air-filled capacitor are 10 cm by 20 cm and the gap between the plates is 6 mm. What is the capacitance when its gap is only half-filled with a dielectric having κ = 3.0? We treat this problem as 2 dielectrics in SERIES

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**Dielectric application - computer key on keyboard**

When the key is pressed, the plate separation is decreased and the capacitance increases. Each key corresponds to a different capacitance.

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**Dielectric application: Stud-finder**

(how you doin!) The dielectric constant of wood (and of all other insulating materials, for that matter) is greater than 1; therefore, the capacitance increases. This increase is sensed by the stud-finder's special circuitry, which causes an indicator on the device to light up.

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