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Here we have an apple that is at a height h above the Earths surface.

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Presentation on theme: "Here we have an apple that is at a height h above the Earths surface."— Presentation transcript:

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2 Here we have an apple that is at a height h above the Earths surface.

3 The apple has a potential energy equal to

4 Anywhere along the dotted line the apple has the same height, therefore the same potential energy.

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7 This line that has the same potential is called an line.

8 This line that has the same potential is called an line.

9 For any object there is an infinite number of lines.

10 Here we show some of the different lines.

11 If we moved the apple along one of these lines

12 there would be no change in potential energy therefore no work would be done.

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31 Once again, there was NO WORK done in moving the apple along the from point A to point B.

32 We also know that if we raise an apple above the Earth and let it go, it will fall.

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38 The direction that the apple accelerates tells us the direction of the gravitational force.

39 The direction of the gravitational force and the pattern of equilpotential lines give us a view of the. By Richard J. Terwilliger

40 around the Earth.

41 means that there will be a gravitational force on an object if it is placed in the field.

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43 Could this model also work with

44 And how would we deal with two different net charges?

45 Lets start with an object that has a NET NEGATIVE charge.

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54 Placing a test charge in the vicinity of this net negative charge and noticing if it experiences a force will tell us if there is an Electric Field around the charge.

55 It will also tell us the direction of the Electric Field.

56 The test charge is always defined as

57 Therefore the test charge has a force acting on it the net negative charge.

58 This is the direction of the

59 Moving the test charge around the net negative

60 and plotting the direction of the force will show us the field surrounding the charge.

61 This positive test charge if free to move will fall towards the net negative charge.

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63 The positive test charge has no potential energy at this point.

64 To pull the positive test charge away from the negative we must do work on the positive test charge.

65 This work is equal to the potential energy at that point.

66 So similar to the equipotentials surrounding the Earth,

67 we have equal potentials surrounding the net negative charge.

68 If the charge moves along the equipotential there is no work done.

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88 By Richard J. Terwilliger

89 If the charge moves along the equipotential there is no work done. By Richard J. Terwilliger

90 If the charge moves along the equipotential there is no work done.

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97 By Richard J. Terwilliger

98 If the charge moves along the equipotential there is no work done.

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101 We now know the direction of the force and the pattern of equipotentials around the net negative charge.

102 Notice the lines of force are at right angles to the equipotential lines. 90 o

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105 We can now predict the electric lines of force and the equipotential lines around a charge.

106 A positive test charge placed near the net positive charge will experience a force outward.

107 Therefore the electric field surrounding the net positive radiates out away from the positive.

108 And the equipotential lines must cross these force lines at right angles forming concentric circles.

109 Notice the force lines never cross each other and the equipotential lines never cross.

110 Copy this diagram into your notebook:

111 So we now know what the fields look like around either a positive or negative charge

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113 What would the electric field lines and the equipotential lines look like around two charges?

114 One negative and one positive.

115 First we place our positive test charge in the field and determine the direction of the force on the test charge.

116 The test charge is repelled away from positive and attracted toward the negative.

117 Now move the test charge to a new position and determine the direction of the force.

118 Keep moving the test charge and determine the direction of the force at each new position.

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120 If we place the test charge at the position shown,

121 The test charge will experience a large force pushing it away from the positive charge and

122 A very small force pulling it towards the negative charge.

123 The electric field, at this point, would be the resultant of these two forces.

124 Move the test charge and again find the resultant.

125 Here the test charge is further away from the positive charge so the force is smaller.

126 Move the test charge and again find the resultant force.

127 Keep repeating until you have the pattern for the electric field line, the line of force.

128 Once more.

129 Connecting all of these arrows gives us the electric field line.

130 From here we can finish the pattern.

131 Now draw in the Remember, they never cross each other and must cross the force lines at right angles.

132 Copy this diagram into your notebook:

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152 What does the Electric Force lines and Equipotential lines look like between parallel plate charges?

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154 Lets check it out!

155 First well start with two parallel plates.

156 Next well charge one plate net negative and the other plate net positive.

157 To determine the direction of the electric field well place the test charge between the plates.

158 Remember the test charge?

159 Is the test charge positive or negative? CLICK on YOUR ANSWER

160 Hello? McFly!

161 The electric field direction is determined using a net test charge

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163 A positive charge placed between the two parallel plates

164 will be forced away from the positive plate and towards the negative plate.

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209 Therefore the electric field direction between two parallel charged plates is away from the positive plate and towards the negative plate.

210 And the equipotential lines are perpendicular to the force lines.

211 Copy this diagram into your notebook:

212 How is electric field strength and electric potential related?

213 V = W/q 2

214 V = W/q 2 V = F * d / q 2

215 V = W/q 2 V = F * d / q 2 V = F/ q 2 * d

216 V = W/q 2 V = F * d / q 2 V = F/ q 2 * d V = E * d Copy this last formula in your notebook:

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