PHConcept Q

# PHConcept Q - 1. 2 aluminized spheres of equal radii hang...

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1. 2 aluminized spheres of equal radii hang by insulating threads. Using a glass rod and a piece of silk, how do you insure the two spheres receive exactly the same charge? A: Rub the rod with the silk and induce charge on the rod. Set up the spheres so both are touching, and touch one sphere with the rod. The same same charge will be transferred to each. Also “charge by induction” would work if spheres are touching. 2. How does the spacing between the equipotential surfaces reflect the strength of the electric field? A: V ~ E (change in r); since V is inversely proportional, decreasing the distance increases the field, and increasing the distance, reduces it. 3. A charged conducting sphere has an electrical potential energy U. What is the potential energy of another conducting sphere with the same charge but of half the radius? A: The potential energy of a sphere is inversely proportional to the radius; thus halving the radius, doubles the potential E. 4. When the separation of the plates of an isolated but charged parallel plate capacitor is increased, what happens to the stored energy in the capacitor? A: U = (½)QV; since the field is constant, V = Ed.U = (½)QEd; charge remains constant, so increasing the distance increases the potential energy stored. 5. What is the electric field inside a spherical insulator of radius R which has charge Q distributed only on its surface? A: If the charge is distributed uniformly, then the net charge within a sphere within the sphere is 0. 6. Sketch the electric field lines and the lines of equipotential around two positive charges. Hint: suppose the charges are separated by a distance d. At a distance far from the charges, how will the lines appear? The lines appear to surround both charges. 7. Two capacitors C1 and C2 are connected either in series or in parallel to a battery providing voltage V. What is the total amount of electrical energy that is stored in the two capacitors in each case? U = (½)CV^2; in a series, Ceq = (1/C1 + 1/C2)^(-1) => U = (½)[(1/C1 + 1/C2)^(- 1)]V^2; in parallel, Ceq = C1 + C2 => U = (½)(C1 +C2)V^2. 8. A very large slab of an insulator has a charge distributed uniformly throughout with a uniform volume charge density p =1.0 C /m^3. The insulating slab is 10 cm thick. A 10 cm think conducting slab is separated by 20 cm from the first slab. The conducting slab is isolated and has zero net charge. 1. Plot the electric field E as a function of x from x<0 to x>40 cm. Give the values (Magnitude and direction) of E at different values of x (x<0, 0<x<10, 10<x<30, 30<x<40, x>40 cm). A: E = pd/2 E o; x<0 => to the left, 0<x<10 => 0, 10<x<30 => to the right, 30<x<40 => 0, x>40 cm => to the right 2. Plot the electric potential V for the same range of x, relative to V = 0 at x = 0.

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## This note was uploaded on 04/18/2008 for the course PHYS 0030 taught by Professor Cutts during the Spring '07 term at Brown.

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PHConcept Q - 1. 2 aluminized spheres of equal radii hang...

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