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    Gauss Law

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    elGauss’ Law AP Physics C Electric Flux Let’s start be defining an area on the surface of an object. The magnitude is “A” and the direction is directed perpendicular to the area like a force normal. A E Flux ( or FLOW) is a general term associated with a FIELD that is bound by a certain AREA. So ELECTRIC FLUX is any AREA that has a ELECTRIC FIELD passing through it. We generally define an AREA vector as one that is perpendicular to the surface of the material. Therefore‚ you can

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    In physics‚ the charge of a volume is commonly described by a quantity called the charge density or the charge distribution. When the electric field that arises from the charge distribution exhibits a volumetric symmetry‚ a handy relationship known as Gauss’s Law may be used to calculate the charge distribution of the volume. The charge volume of a ball or rod mill is expressed as the percentage of the volume within the liners filled with balls or rods. When the mill is stationary‚ the charge

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    Electric Charge

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    Part 1: What is the relationship between the electric potential at a point in space and the distance from an electric charge? 1)      Place a positive 1 nano-coulomb charge on the screen. 2)      Turn on “Show numbers” 3)      Turn on “tape measure”. 4)      Use the tape measure to find and record the distance from the charge to the equipotential sensor. 1.88 m 5)      Record the voltage as indicated on the equipotential sensor. 4.1v 6)      Change the location of the positive charge to

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    Phyc 260 Solutions

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    Chapter 24 Gauss’s Law Multiple Choice 1. Two charges of 15 pC and –40 pC are inside a cube with sides that are of 0.40-m length. Determine the net electric flux through the surface of the cube. a. +2.8 N ( m2/C b. –1.1 N ( m2/C c. +1.1 N ( m2/C d. –2.8 N ( m2/C e. –0.47 N ( m2/C 2. The total electric flux through a closed cylindrical (length = 1.2 m‚ diameter = 0.20 m) surface is equal to –5.0 N ( m2/C. Determine the net charge within the cylinder

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    Morgan Land Elizabeth Hollis Physics II February 7‚ 2012 Electric Fields Lab Abstract: In this lab we will study the equipotential lines in an electric field in order to study the structure of the electric lines of force. We will plot the position of electric field lines in a given electric field using both a manual method and a computer software program. There is an electric field in any

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    Lab: Electrostatic Coulomb Constant Rylee Rose Date: 1/28/13 Abstract: This experiment was done to study electrostatic and Coulomb’s law. Objective: The objective was to demonstrate the veracity of Coulomb’s Law. This was done by observing different equilibrium conditions of two pith balls that were equally charged. They were then constructed into a pendulum using strings. Calculations were done to determine the charge on pith balls‚ calculate the mass of the pith balls and to calculate

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    Electrostatics Lab Report

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    Electrostatics Lab Name: Bryan Michel Materials: Styrofoam cups Styrofoam plates PVC Pipe Fur Scotch Tape Procedures: You will be investigating charging object by friction. When you are asked to write your observations‚ please do so in a clear‚ complete manner. 1. Tape a piece of string to a Styrofoam cup and hang it from the edge of the desk so that it can move freely. Rub the outside of the cup with fur. Rub a PVC pipe with fur and bring it near the hanging cup.

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    Phil Yu Electrostatics Lab Data Chart Part 1: Trial | Material | Material | Charge (+ or -) | q (nC) | 1 | Glass | Fur | + | 3.108 | 2 | Clear Plastic | Fur | + | 45.07 | 3 | Yellow Plastic | Fur | - | 81.57 | 4 | Glass | Silk | + | 1.979 | 5 | Clear Plastic | Silk | + | 19.60 | 6 | Yellow Plastic | Silk | - | 75.49 | 7 | Glass | Green Cloth | + | 2.914 | 8 | Clear Plastic | Green Cloth | + | 31.09 | 9 | Yellow Plastic | Green Cloth | - | 90.30 |     Data Chart Part 2:

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    Electrostatics Test Paper

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    CBSE TEST PAPER-01 CLASS - XII PHYSICS (Unit – Electrostatics) [ANSWERS] FV force between the charges in vaccuum = FM force between two charges in medium FV ⇒ Fm = k ⇒ if k increases‚ Fm decreases. Ans 01. Since K= Ans 02. Suppose rod P be negatively charged since it attracts rod R ⇒ R is positively charged since it repels rod Q ⇒ Q is negatively charged. So force between Q and R is attractive in nature. Ans 03. Since F = q E and a = F/m Since charge on proton

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    Results and Discussion As there were five similar experiments done‚ there is one common observed behavior – similar charges repel and opposite charges attract. The experiment went expected as there were no unusual observations recorded. In the first part‚ it was observed that when the plastic rod was rubbed it attracted the tissue bits. The explanation with this is‚ when the plastic rod was rubbed‚ it became electrically charged. Conduction happens when the tissue paper was rubbed into the plastic

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