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0 10 10 c r fp 64 10 5 n x question 3 solution

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Unformatted text preview: s case, the negative solution refers to a point to the right of the negative charge. This puts our field point between the charges, which is a scenario we’ve already dismissed. Therefore, our field point P must be located 1.8 m to the left of the negative charge. Solutions will be posted at www.prep101.com/solutions 8 ©Prep101 Phys1026 April Exam Solutions Question 2: Solution: For A: r r r k (Q1 - Q 2 ) 8.99 × 10 9 Nm 2 C -2 × 4.0 × 10 −7 C ˆ ˆ E p = E p1 + E p2 = x= x ( d )2 ( 0.15 m )2 2 2 r 5 -1 ˆ E p = 6.39 × 10 x N ⋅ C (or V ⋅ m -1 ) ( )( ) For B: r r ˆ Fp = qE p = 6.4 × 10 5 N ⋅ C −1 x × 1.0 × 10 −10 C r ˆ Fp = 6.4 × 10 -5 N x ( )( ) Question 3: Solution: Answer (e) Since B is repelled by A, we know that ball B is positively charged. However, C may be attracted to B either because it is negatively charged, or because of an induced charge making its left side slightly more negative and its right side slightly more positive. Question 4: Solution: After simplification gives: Solutions will be posted at www.prep101.com/solutions 9 ©Prep101 Phys1026 April Exam Solutions Question 5: Solution: Consider a differential element of length d l = Rd θ , which makes an angle θ with the xaxis, as shown in the figure. The amount of charge it carries is dq = λ d l = λ Rd θ The contribution to the electric field at O is: Integrating over the angle from − θ o to + θ o , we have We see that the electric field only has the x-component, as required by a symmetry argument. If we take the limit θ o → π , the arc becomes a circular ring. Since sin π = 0 , the equation above implies that the electric field at the center of a non-conducting ring is zero. This is to be expected from symmetry arguments. On the other hand, for very small θ o , sin θ o ≈ θ o and we recover the point-charge limit: where the total charge on the arc is Q = λ l = λ (2 R θ o ) Answer 3.6 The net electric field at point D has components from the charges at A, B, C: k (−Q) kQ EB = = 2 N/C [DOWN] d2 d k (−Q) kQ EC = = 2 N/C [LEFT] d2 d k (Q) kQ N/C = EA = 2 2 2 2d 2 d +d ( ) Solutions will be posted at www.prep101.com/solutions 10 ©Prep101 Phys1026 April Exam Solutions kQ kQ cos 45° N/C; E Ay = sin 45° N/C 2 2d 2d 2 The net electric field is the sum of the components: kQ kQ kQ ⎛ cos 45° ⎞ kQ E at D , x = cos 45° − 2 = 2 ⎜ − 1⎟ = −0.65 2 N/C 2 d d⎝ 2 d 2d ⎠ kQ kQ kQ ⎛ sin 45° ⎞ kQ E at D , y = sin 45° − 2 = 2 ⎜ − 1⎟ = −0.65 2 N/C 2 d d⎝ 2 d 2d ⎠ kQ kQ ⎞ ⎛ E at D = ⎜ − 0.65 2 ,−0.65 2 ⎟ N/C d d⎠ ⎝ kQ kQ ⎞ ⎛ Field at D = E at D = ⎜ − 0.65 2 ,−0.65 2 ⎟ N/C d d⎠ ⎝ E Ax = GAUSS’ LAW PRACTICE QUESTIONS Question 1: Solution: Recall that the flux is given as Φ E = E ⋅ A = EA cos θ where θ is the angle between the vector area A and the electric field E. In order for the flux to be at a maximum, we must have cos θ = 1 , which will occur when the vector area of the loop is aligned parallel to the field. In the case of...
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