# C to nd constant at t 0 z t 0 d 3 d 2 tan1 3 d

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Unformatted text preview: 1 d − 3 − z d(d − z ) + d 2 tan−1 z = −t d−z q2 + 8 π ε0 m C constant We use I.C. to ﬁnd constant ⇒ at t = 0, z (t = 0) = d. 3 ⇒ d 2 tan−1 π3 d = C ⇒ C = d2 0 2 π 2 plugging in z = 0 gives the time (T) it takes for the charge to reach the z = 0 ground plane. 3 ⇒ 0 + d 2 tan−1 0 =− d q2 π3 T + d2 8 π ε0 m 2 0 2 Spring 2006 Quiz 1 π3 2 2d ⇒T = q2 8π ε0 m 6.641, Spring 2005 = π 2 3 8 π ε0 m d ⇒ T= 4 q2 2π 3 d3 ε0 m q2 Problem 2 y λ ε0 + + + + II -a I + a +++++++ -L + a φ +++++++ III L Figure 2: Uniformly distributed line charge λ. (Image by MIT OpenCourseWare.) Φ(r ) = l′ λ(r ′ )dl′ 4π ε0 |r − r ′ | λ(r ′ )ir′ r dl′ E (r ) = l′ 2 4π ε0 |r − r ′ | A Question: Find the p otential at the p oint (x = 0, y = 0, z = 0) −a Φ(x = 0, y = 0, z = 0) = x =− L λdx + 4π ε0 (−x) π φ=0 λadφ + 4 π ε0 a −a π λ...
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