PHY 184 Spring 2007 Lecture 14 1/31/07 184 Lecture 14

Announcements Midterm 1 will take place in class a week from tomorrow, Thursday, February 8. One 8.5 x 11 inch equation sheet (front and back) is allowed The exam will cover Chapters 16 - 19 Homework Sets 1 - 4 Practice exam (the actual Midterm 1 from last fall) is available: http://www.pa.msu.edu/courses/phy184 1/31/07 184 Lecture 14

Review The electric potential energy stored in a capacitor is given by The field energy density stored in a parallel plate capacitor is given by The field energy density in general is 1/31/07 184 Lecture 14

Review (2) Placing a dielectric between the plates of a capacitor increases the capacitance by  (dielectric constant) The dielectric has the effect of lowering the electric field between the plates (for given charge q) We also define the electric permitivity of the dielectric material as 1/31/07 184 Lecture 14

Dielectric Strength The “dielectric strength” of a material measures the ability of that material to withstand voltage differences. If the voltage across a dielectric exceeds the breakdown potential, the dielectric will break down and begin to conduct charge between the plates. Real-life dielectrics enable a capacitor provide a given capacitance and hold the required voltage without breaking down. Capacitors are usually specified in terms of their capacitance and rated voltage. 1/31/07 184 Lecture 14

Dielectric Constant The dielectric constant of vacuum is defined to be 1. The dielectric constant of air is close to 1 and we will use the dielectric constant of air as 1 in our problems. The dielectric constants of common materials are 1/31/07 184 Lecture 14

Capacitor with Dielectric Question 1: Consider a parallel plate capacitor with capacitance C = 2.00 F connected to a battery with voltage V = 12.0 V as shown. What is the charge stored in the capacitor? Question 2: Now insert a dielectric with dielectric constant  = 2.5 between the plates of the capacitor. What is the charge on the capacitor? The additional charge is provided by the battery. 1/31/07 184 Lecture 14

Capacitor with Dielectric (2) Question 3: We isolate the charged capacitor with a dielectric by disconnecting it from the battery. We remove the dielectric, keeping the capacitor isolated. What happens to the charge and voltage on the capacitor? The charge on the isolated capacitor cannot change because there is nowhere for the charge to flow. Q remains constant. The voltage on the capacitor will be The voltage went up because removing the dielectric increased the electric field and the resulting potential difference between the plates. V increases 1/31/07 184 Lecture 14

Capacitor with Dielectric (3) Question 4: Does removing the dielectric from the isolated capacitor change the energy stored in the capacitor? The energy stored in the capacitor before the dielectric was removed was After the dielectric is removed, the energy is The energy increases --- we must add energy to pull out the dielectric. [Or, the polarized dielectric is sucked into the E.] 1/31/07 184 Lecture 14

Example A coaxial cable used in a transmission line has an inner radius of a=0.10 mm and an outer radius of b=0.60 mm. Calculate the capacitance per meter for this type of cable. Idea: The capacitance of a cylindrical capacitor is given by The capacitance per unit length is thus: 1/31/07 184 Lecture 14

Clicker Question A coaxial cable used in a transmission line has an inner radius of a=0.10 mm and an outer radius of b=0.60 mm. We calculate the capacitance per meter to be 81 pF/m. Now assume that the space between the conductors is filled with material of =3. How does the capacitance change? A) 243 pF/m B) 27 pF/m C) doesn’t change D) 8.1 pF/m 1/31/07 184 Lecture 14

Example An air-filled parallel plate capacitor has a capacitance of 1.3 pF. The separation of the plates is doubled, and wax is inserted between them. The new capacitance is 2.6pF. Find the dielectric constant of the wax. Key Ideas: The original capacitance is given by Then the new capacitance is Thus rearrange the equation: 1/31/07 184 Lecture 14

Example Given a 7.4 pF air-filled capacitor. You are asked to convert it to a capacitor that can store up to 7.4 J with a maximum voltage of 652 V. What dielectric constant should the material have that you insert to achieve these requirements? Key Idea: The capacitance with the dielectric in place is given by C=Cair and the energy stored is given by So, 1/31/07 184 Lecture 14

Clicker Question - part 1 A parallel-plate air-filled capacitor has a capacitance of 50 pF. (a) If each of the plates has an area of A=0.35 m2, what is the separation? A) 12.5 10-1 m B) 6.2 10-2 m C) 1.3 m 0=8.85 10-12 C2/Nm2 1/31/07 184 Lecture 14

Clicker Question - part 1 A parallel-plate air-filled capacitor has a capacitance of 50 pF. (a) If each of the plates has an area of A=0.35 m2, what is the separation? B) 6.2 10-2 m Use to solve for d: 1/31/07 184 Lecture 14

Clicker Question - part 2 An air-filled parallel plate capacitor has a capacitance of 50pF. (b) If the region between the plates is now filled with material having a dielectric constant of =2, what is the capacitance? A) the same B) 25 pF C) 100 pF 1/31/07 184 Lecture 14

Clicker Question - part 2 A air-filled parallel plate capacitor has a capacitance of 50 pF. (b) If the region between the plates is now filled with material having a dielectric constant of =2, what is the capacitance? C) 100 pF 1/31/07 184 Lecture 14

Microscopic Perspective on Dielectrics Let’s consider what happens at the atomic and molecular level when a dielectric is placed in an electric field. There are two types of dielectric materials Polar dielectric Non-polar dielectric Polar dielectric material is composed of molecules that have a permanent electric dipole moment due to their molecular structure e.g., water molecules Normally the directions of the electric dipoles are randomly distributed as shown below 1/31/07 184 Lecture 14

Microscopic Perspective on Dielectrics (2) When an electric field is applied to these polar molecules, they tend to align with the electric field Non-polar dielectric material is composed of atoms or molecules that have no electric dipole moment 1/31/07 184 Lecture 14

Microscopic Perspective on Dielectrics (3) These atoms or molecules can be induced to have a dipole moment under the influence of an external electric field. This induction is caused by the opposite direction of the electric force on the negative and positive charges of the atom or molecule, which displaces the center of the relative charge distributions and produces an induced electric dipole moment 1/31/07 184 Lecture 14

Microscopic Perspective on Dielectrics (4) In both the case of the polar and non-polar dielectric materials, the resulting aligned electric dipole moments tend to partially cancel the original electric field The electric field inside the capacitor then is the original field minus the induced field 1/31/07 184 Lecture 14

Example: Camera Flash 1/31/07 184 Lecture 14

Example: Camera Flash 1/31/07 184 Lecture 14

Example: Camera Flash 1/31/07 184 Lecture 14

Partially Filled Capacitor A cylindrical capacitor is lying on a table as shown. The outer cylinder has radius b=4.5 cm, and the inner cylinder has a radius a=4.0 cm. The total length of the capacitor is L=9.0 cm. A dielectric material of constant κ=3.5 is inserted a length L-d=3 cm between the two conducting cylinders. The partially filled capacitor is then connected to a voltage 100.0 V. In what follows, we'll refer to the region where there is no dielectric as region I and the region where there is a dielectric present as region II. What is the ratio of the charge in region I to that in region II? 1/31/07 184 Lecture 14

Partially Filled Capacitor - 2 The voltage between the two cylinders is V, regardless of whether we are in region I or II (it is like two capacitors in parallel); since q = CV, where the common factor is cancelled The partially filled capacitor is disconnected from the voltage source, so that it remains charged. The dielectric is then moved further into the capacitor, so that region II now has length L-d' = 3.5 cm. Now what is the voltage difference between the cylinders? Here the total charge is what is was before, so the new voltage is 1/31/07 184 Lecture 14

Partially Filled Capacitor - 3 Is a force required to hold the dielectric in place, and if so, what is its direction? The energy of a capacitor is qV/2; since V was reduced while q remained constant, the energy of the capacitor is lowered by inserting the dielectric into it. This means that the work was done by the capacitor+dielectric, ie, if we attached the left-hand side of the dielectric (the side outside the capacitor) to a spring the dielectric would pull on the spring. A force is required to keep the dielectric from moving further into the capacitor. 1/31/07 184 Lecture 14