Eavg = v/d V=kE/r V=-b integration E.dl A uniformly charged, thin ring has radiu

Question

Eavg = v/d V=kE/r V=-b integration E.dl A uniformly charged, thin ring has radius of 15.0 cm, total charge +24.0 nC. An electron is placed at the ring's axis a distance of 30.0 cm from center of ring is constrained to stay on the axis of ring. Find the speed of the electron when it reaches the center of ring A solid conducting where has a net positive charged radius of R=0.4 cm. At a point 1.20 m from center of sphere, the electron potential is measured to be 24.0V. What is the electric potential at the center of sphere? Give explanation.

Explanation / Answer

part 1 )

Apply conservation of energy to the motion of the electron.

V = kQ/sqrt(x^2 + R^2)

xa = 30 cm , xb = 0

KEa = 0 (released from rest ) , KEb = 1/2mv^2

1/2*mv^2 = Ua - Ub

U = qV = -eV

v = sqrt(2e(Vb - Va)/m)

Va = kq/sqrt(xa^2 + R^2) = 9 x 10^9 x 24 x 10^-9 / sqrt[(0.300)^2 + (0.150)^2 ]

Va = 643 V

Vb = kQ/R = 9*10^9 x 24 x 10^-9 / 0.150 = 1438 V

v = sqrt(2*(1.6x10^-19)(1438 - 643) / 9.1 x 10^-31) = 1.67 x 10^7 m/s

part 2 )

V = kQ/r

Q = V*r/k

Q =24 * 1.2 / 9 x 10^9

Q = 3.2 x 10^-9 C

at center

V = kQ/R

V = 9 x 10^9 x 3.2 x 10^-9 /0.400

V = 72 V

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