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Chapter 21- Examples 1 Problem Two free point charges +q and +4q are located a distance, L, apart. A third charge is placed so that the whole system is in equilibrium Find the location, magnitude, and sign of the third charge 2 Step 1: Draw IT! +q L +4q 3 Step 2: Think about it! 1 +q 2 L +4q 3 There are 3 positions available for the new charge (marked in blue). Which position do you think is the most likely position and more importantly why? Positions 1 & 3 are untenable since by equilibrium, we mean the force on each particle is zero. If 1, for example, then there will be a repulsive force on +4q that cannot be overcome without creating a strong attractive force on +q 4 Step 3: Now a free body problem +q 2 +4q L x F1 F2 Call the new charge, q0, and let it be distance x from +q The free-body diagram shows relationship of the two forces, F1 and F2 on the new charge qq0 4qq0 k 2 k x ( L x) 2 1 4 2 x ( L x) 2 ( L x) 2 4 x 2 L x 2x L x 3 5 Step 4: Use the other equilibrium points to derive more information F4 +q 2 +4q L x F1 The free-body diagram shows relationship of the two forces, F1 and F4 on charge +q qq0 4q q k 0 where x2 ( L) 2 q0 4 q 2 ( L) 2 L 9 9 q 0 4 q k L x 3 4 q0 q 9 6 Problem An electron is projected with an initial speed of v0, 1.6 x 106 m/s in the uniform field between the parallel plates shown below. Assume that the plates are uniform and the field is directed vertically downward and the field outside the plates is zero. The electron enters the field at a point midway between the two plates If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field 2.0 cm 1.0 cm e- 7 Step 1: Draw It Obviously, the electron is 0.5 cm below the upper plate when entering the field so, the initial position is (0,0.5) and the final position is (2.0, 0.0) SI: (0,0.005) and (0.02, 0) Also, v0 is actually in the x-direction so vinitial= (v0,0) 2.0 cm 1.0 cm e- 8 Step 2: Think about it! This reminds me of a projectile problem except it is kinda upside down. In this case, there is no acceleration in the x-direction only in the y and since F=ma=eE then a=eE/m Where e=charge of electron and m is the mass of the electron. Since I can never remember the trajectory equation, then I must solve for the time it takes the electron to traverse 2 cm and then plug that into the an equation relating acceleration to distance in the y-direction 2.0 cm 1.0 cm e- 9 Step 3: Equation Time 2.0 cm 1.0 cm e- x v0t x0 0.02 (1.6 106 )t 0 t 1.25108 s 1 2 at y0 2 y y0 .005m y 0.010 a(1.25108 s) 2 a 6.4 1013 m / s 2 m 9.11 1031 kg and e 1.602 1019 C eE ma a or E m e (9.11 1031 )(6.4 1013 ) E 19 1.60210 N 364 E C 10 Problem What is the magnitude and direction of the electric field at the center of the square if q= 1.0 x 10-8 C and distance, a, is 5 cm +q a a -q -2q a a +2q 11 Field Lines +q a -2q Free body diagram of + test charge in center 2q a -2q a -q q -q a +2q 12 Need break down into x and y components First find the magnitudes of all the electric fields 2q E2 q k 2 r q Eq k 2 r 2 2 a2 1 1 2 where r a a 2 2 2 13 For example, the field line from +2q The line from +2q makes an angle of 135 1350 with the positive x-axis So E2qx=E2q cos (1350)=-E2q cos (450) 0 E2qx=-0.707*E2q E2qy=E2q sin (1350)=E2q sin (450) E2qy=0.707*E2q 14 And from -2q, (at 450) E-2qx=0.707*E2q E-2qy=0.707*E2q 15 And –q and +q -q makes angle of 2250 E-qx=-0.707*Eq E-qy=-0.707*Eq -q makes angle of 3150 Eqx=0.707*Eq Eqy=-0.707*Eq Now we add the x and y components, respectively 16 Ex and Ey 2q E2 q k 2 r q Eq k 2 r a2 where r 2 E2 q 2 Eq k Eq k 2 E x E2 qx E 2 qx Eqx E qx E x 0.707* ( E2 q E2 q Eq Eq ) Ex 0 E y E2 qy E 2 qy Eqy E qy E y 0.707* ( E2 q E2 q Eq Eq ) 4q a2 E y 0.707* (2 Eq 2 Eq Eq Eq ) E y 2 Eq 2 k 2q a2 2q a2 2q E E x2 E y2 2k 2 a tan Ey Ex Undefined tan1 (undefined) 900 17