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PHY 184

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PHY 184

Spring 2007

Lecture 12

Title: Capacitor calculations

184 Lecture 12

- Homework Set 3 is due tomorrow morning at 8:00 am.
- Midterm 1 will take place in class next week on Thursday, February 8.
- Practice exam will be posted in a few days.
- Second half of this Thursday’s lecture: review.

184 Lecture 12

- The definition of capacitance is
- The unit of capacitance is the farad (F)
- The capacitance of a parallel plate capacitor is given by
- Variables:
A is the area of each plate

d is the distance between the plates

184 Lecture 12

- Consider a capacitor constructed of two collinear conducting cylinders of length L.
- The inner cylinder has radius r1 andthe outer cylinder has radius r2.
- Both cylinders have charge perunit length with the inner cylinderhaving positive charge and the outercylinder having negative charge.
- We will assume an ideal cylindrical capacitor
- The electric field points radially from the inner cylinder to the outer cylinder.
- The electric field is zero outside the collinear cylinders.

184 Lecture 12

- We apply Gauss’ Law to get the electric field between the two cylinder using a Gaussian surface with radius r and length L as illustrated by the red lines
- … which we can rewrite to get anexpression for the electric fieldbetween the two cylinders

184 Lecture 12

- As we did for the parallel plate capacitor, we define the voltage difference across the two cylinders to be V=V1 – V2.
- The capacitance of a cylindrical capacitor is

Note that C depends on geometrical factors.

184 Lecture 12

- Consider a spherical capacitor formed by two concentric conducting spheres with radii r1 and r2

184 Lecture 12

- Let’s assume that the inner sphere has charge +q and the outer sphere has charge –q.
- The electric field is perpendicular to the surface of both spheres and points radially outward

184 Lecture 12

- To calculate the electric field, we use a Gaussian surfaceconsisting of a concentric sphere of radius r such that r1 < r < r2
- The electric field is always perpendicular to the Gaussian surface so
- … which reduces to

…makes sense!

184 Lecture 12

- To get the electric potential we follow a method similar to the one we used for the cylindrical capacitor and integrate from the negatively charged sphere to the positively charged sphere
- Using the definition of capacitance we find
- The capacitance of a spherical capacitor is then

184 Lecture 12

- We obtain the capacitance of a single conducting sphere by taking our result for a spherical capacitor and moving the outer spherical conductor infinitely far away.
- Using our result for a spherical capacitor…
- …with r2 = and r1 = R we find

…meaning V = q/4pe0R

(we already knew that!)

184 Lecture 12

b=40 mm

a=38 mm

d

A ?

- The “plates” of a spherical capacitor have radii 38 mm and 40 mm.
a) Calculate the capacitance.

b) Calculate the area A of a parallel-plate capacitor with the same plate separation and capacitance.

Answers: (a) 84.5 pF; (b) 191 cm2

184 Lecture 12

- Two metal objects have charges of 70pC and -70pC, resulting in a potential difference (voltage) of 20 V between them. What is the capacitance of the system?

A) 140 pF

B) 3.5 pF

C) 7 pF

D) 0

184 Lecture 12

The capacitance is the constant of proportionality between change and voltage. It depends on the geometry not on the charge or voltage.

- Two metal objects have charges of 70pC and -70pC, resulting in a potential difference (voltage) of 20 V between them. How does the capacitance C change if we double the charge on each object?

A) C doubles

B) C is cut in half

C) C does not change

184 Lecture 12

- A circuit is a set of electrical devices connected with conducting wires.
- Capacitors can be wired together in circuits in parallel or series
- Capacitors in circuits connected by wires such that the positively charged plates are connected together and the negatively charged plates are connected together, are connected in parallel.
- Capacitors wired together such that the positively charged plate of one capacitor is connected to the negatively charged plate of the next capacitor are connected in series.

184 Lecture 12

.. key point for capacitors in parallel

- Consider an electrical circuit with three capacitors wired in parallel
- Each of three capacitors has one plate connected to the positive terminal of a battery with voltage V and one plate connected to the negative terminal.
- The potential difference V across each capacitor is the same.
- We can write the charge on each capacitor as …

184 Lecture 12

- We can consider the three capacitors as one equivalent capacitor Ceq that holds a total charge q given by
- We can now define Ceq by
- A general result for n capacitors in parallel is
- If we can identify capacitors in a circuit that are wired in parallel, we can replace them with an equivalent capacitance

184 Lecture 12

.. key point for capacitors in series

- Consider a circuit with three capacitors wired in series
- The positively charged plate of C1 is connected to thepositive terminal of the battery.
- The negatively charge plate of C1 is connected to thepositively charged plate of C2.
- The negatively charged plate of C2 is connected to thepositively charge plate of C3.
- The negatively charge plate of C3 is connected to thenegative terminal of the battery.
- The battery produces an equal charge q on each capacitor because the battery induces a positive charge on the positive place of C1, which induces a negative charge on the opposite plate of C1, which induces a positive charge on C2, etc.

184 Lecture 12

- Knowing that the charge is the same on all three capacitorswe can write
- We can express an equivalent capacitance Ceq as
- We can generalize to n capacitors in series
- If we can identify capacitors in a circuit that are wired in series, we can replace them with an equivalent capacitance.

184 Lecture 12

Ceq = 10 pF

Answer: 90 pC

- C1=C2=C3=30 pF are placed in series. A battery supplies 9 V. What is the charge q on each capacitor?

A) q=90 pC

B) q=1 pC

C) q=3 pC

D) q=180 pC

184 Lecture 12

Ceq = 3 pF

Answer: 30 volts

- C1=C2=C3=1 pF are placed in parallel. What is the voltage of the battery if the total charge of the capacitor arrangement, q1+q2+q3, is 90 pC?

A) 180 V

B) 10 V

C) 9 V

D) 30 V

184 Lecture 12