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Three charges are place at the corners of a 2m X 2m square as shown below. What charge must be placed at the fourth corner so that the electric potential is zero at the center of the square (point O)? How much work would be required to bring the fourth charge from infinity and place it at its corner of the square? q 3 30C q4 ? 2m q1 20C O 2m q 2 40C O O,i O,1 O,2 O,3 O,4 0 i O,4 O,1 O,2 O,3 kq i i r O,i kq1 kq 2 kq 3 kq 4 rO,4 r r r O,1 O,2 O,3 rO,1 rO,2 rO,3 rO,4 rO k q k q q q 2 3 rO 4 rO 1 q 4 q 1 q 2 q 3 q 4 20C 40C 30C q 4 10C Finding the potential at the fourth corner prior to placing q 4 there : 4 4,i 4,1 4,2 4,3 i kq i 4,i r 4,i kq1 kq 2 kq 3 4 r r r 4,1 4,2 4 ,3 q 1 q2 q 3 4 kr r r 4,1 4,2 4,3 r4,1 r4,3 2m r4,2 2m 2 2.83m 2 6 6 6 Nm 20 10 C 40 10 C 30 10 C 4 9 10 2m 2.83m 2m m2 9 4 9.78 10 volts 4 Wq 4 q 4 4 4 Wq 4 30 10 C9.78 10 volts 0 6 4 4 Wq 4 2.93J 4 No work is required by an external force to bring q 4 from infinity and place it at the fourth corner...The electric field does the work. A capacitor is oriented so that its plates lie horizontally, negative plate below the positive plate. The plates carry charges of +/-20C, are separated by a distance of 4m and have an area of 0.1m2. An object with a mass of 1.5kg and carrying a charge of 5C is ejected vertically upward from the negative plate with an initial velocity of 25m/s. A) Will it reach the upper plate? If not how far from the upper plate will it stop? B) If the capacitor were rotated so that its plates were vertical, would the object reach the positive plate? If it reaches the positive plate what will be its velocity? A 0.1m 2 Part A 20C ? 4m v i 25 m s 20C m 1.5kg, q 5C As the object moves between the plates of the capacitor the only forces acting on it are the gravitational force and the electric force, both of which are conservative. Therefore, during its motion the total mechanical energy of the object remains constant. If we choose the negative plate as the reference level for both the gravitational and electric potential energies, the initial energy of the object is totally kinetic. As it rises the kinetic energy is converted into gravitational and electrical potential energy. This will continue until either all of the kinetic energy has been converted to potential energy or the object reaches the upper positive plate. If it does not reach the upper positive plate, there is a point where the total mechanical energy is all potential. In either case the total mechanical energy of the object remains constant. Ei E f 1 mv2 1 mv2 mgh q i f f f 2 2 q hf f A o We can find the value of hf where the object stops, vf = 0. If this is greater than the distance between the plates (4m) the object will reach the upper plate. =0 1 mv2 1 mv2 mgh q i f f f 2 2 1 mv2 mgh q i f f 2 1 mv 2 mgh qq h f i f 2 oA qq 1 mv2 mg h i 2 o A f 1 mv2 i 2 h f qq mg o A 2 m .51.5kg 25 s hf 6 6 5 10 C 20 10 C m 1.5kg 9.8 2 2 s 12 C 2 8.85 10 0.1m 2 Nm hf 3.67m The object does NOT reach the upper plate. It stops 0.33m from the positive plate Part B 20C m 1.5kg, q 5C 20C ? v i 25 m s 4m A 0.1m 2 20C q 20C Electric Field vi FE FG vy Gravitational Field vx As the object moves to the right two things happen: 1. Horizontally (FE is opposite to vx) Kinetic energy is converted into electrical potential energy. 2. Vertically (FG is in the same direction as vy) Gravitational potential energy is converted into kinetic energy. Since both forces are conservative the total mechanical energy will remain constant. Horizontal Motion Ei,x Ef ,x Ignore Gravitational Potential Energy 1 mv2 q 1 mv2 q i,x E ,i f ,x E,f 2 2 1 mv q 1 mv2 q qx f ,x 2 2 oA 2 i,x 0 Where x equals object' s horizontal displacement. 1 mv2 1 mv0 2 q q x i,x f ,x 2 2 oA If when vx = 0 x > 4m the object will reach the positive plate. 1 mv2 q q x i,x 2 oA 2 o A 1 x 2 mvi,x qq 2 12 2 C 2 8.85 10 2 0.1m 1 m x 1.5kg 25 Nm 2 s 6 6 5 10 C 20 10 C x 4.15m 4m The object reaches the positive plate. At the positive plate: 1 mv2 1 mv2 q q d i,x f ,x 2 2 oA Where d equals the distance between the plates. vf ,x v f ,x qq d mv 2 A o m 2 i ,x 6 6 2 5 10 C20 10 C 4m m 1.5kg25 s 2 2 12 2 C 8.85 10 2 0.1m N m 1.5kg vf ,x 4.73 m s Time required to reach the positive plate: m 4.73 m 25 vi v f s s 14.9 m v= 2 2 s 4m .27s t d v 14.9 m s Vertical Motion 0 v f,y v i,y gt vf,y v f,y gt m 9.8 2 .27s s vf,y 2.65 m s E i,y E f ,y Ignore Electric Potential Energy 1 mv02 mgh0 1 mv2 mgh i,y i f ,y f 2 2 2 1 mghf mvf ,y 2 2 v f ,y hf 2g 2 m 2.65 s hf m 29.8 2 s hf .36m Final Velocity vf,x 4.73 m s vf,y 2.65 m s v f v 2f ,x v 2f ,y 2 2 m m vf 4.73 s 2.65 s vf 5.42 m s 360 tan 1 v f,y vf,x 2.65 m 1 s 360 tan 4.73 m s 331 vf 5.42 m s @ 331