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Unformatted text preview: I. SAMPLE EXAM PROBLEM 1. A. The electric field is just given by:→ E =→ E 4 Q +→ E Q = 4 Qk a 2 ˆ x + Qk a 2 ˆ x = 5 Qk a 2 ˆ x B. The force is then just:→ F = q o→ E = 5 q o Qk a 2 ˆ x C. Again the electric field is just the electric field due to the 2 charges. So we get, using that the :→ E =→ E 4 Q +→ E Q = 4 Qk 2 a 2 (cos 45ˆ x + sin 45ˆ y ) + Qk 2 a 2 (cos 45ˆ x sin 45ˆ y ) = 2 . 5 Qk √ 2 a 2 ˆ x + 1 . 5 Qk √ 2 a 2 ˆ y D. The plot of the electric field lines is presented in the Figure 1. 2 FIG. 1: Electric Field lines E. At a positions x > a we have that the distance between the negative charge and the point is d = x a , while the distance to the positive charge is just D = x + a . Therefore we get:→ E =→ E 4 Q +→ E Q = 4 Qk ( x + a ) 2 ˆ x Qk ( x a ) 2 ˆ x F. The position x where the field is zero, has to be to the right of the negative charge, since in the middle, both fields are pointing in the same direction, while to the left of the negative charge, the field will never be zero since we will always be closer to the strongest of the 2 charges. Therefore, we just need to set out answer to Section I E to zero. 3 4 Qk ( x + a ) 2 ˆ x Qk ( x a ) 2 ˆ x = 0 4 Qk ( x a ) 2 = Qk ( x + a ) 2 4( x 2 + a 2 2 ax ) = x 2 + a 2 + 2 ax 3 x 2 + 3 a 2 10 ax = 0 x = a 3 x = 3 a where we chose the second solution because it was the only one with x > a (i.e. the other one does not make sense). G. Just like for electric field, we just add the potential of the 2 charges: V = 4 Qk a Qk a = 3 Qk a H. The potential at point P is then just: V = 4 Qk √ 2 a Qk √ 2 a = 3 Qk √ 2 a I. This is a little tricky, but here we go. For any point ( x,y ) with x > 0 since the point has to be closer to the negative charge in order to cancel the ”larger” positive charge we have: V = 4 Qk p ( x + a ) 2 + y 2 Qk p ( x a ) 2 + y 2 Setting the previous result equal to zero we get: 4 4 Qk p ( x + a ) 2 + y 2 Qk p ( x a ) 2 + y 2 = 0 4 Qk p ( x a ) 2 + y 2 = Qk p ( x + a ) 2 + y 2 16( x 2 + a 2 2 ax + y 2 ) = x 2 + a 2 + 2 ax + y 2 15 x 2 + 15 a 2 + 15 y 2 34 ax = 0 x 2 + a 2 + y 2 34 15 ax = 0 x 17 15 a 2 8 a 15 2 + y 2 = 0 x 17 a 15 2 + y 2 = 8 a 15 2 This is the equation of a circle centered at ( 17 a 15 , 0) with a radius R = 8 a 15 . Any point on that circle will have a potential V = 0. J. The energy stored in the system is just: U = q 1 q 2 k r 12 = 4 Q 2 k 2 a = 2 Q 2 k a (1) II. SAMPLE EXAM PROBLEM 3. A. This is almost trivial. There is no electric field inside of a conductor, therefore→ E = 0 , meaning by Gauss’s law that the charge must lie on the surface of the sphere. B. To find the electric field in between the sphere and the shell, we just use Gauss’s law....
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 Spring '10
 GRUNER
 Physics, Charge, Force, Electric charge

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