P24_029 - yields qE − T sin θ = 0 and the sum of the...

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29. The forces acting on the ball are shown in the diagram below. The gravitational force has magnitude mg ,whe re m is the mass of the ball; the electrical force has magnitude qE ,whe re q is the charge on the ball and E is the magnitude of the electric Feld at the position of the ball; and, the tension in the thread is denoted by T . The electric Feld produced by the plate is normal to the plate and points to the right. Since the ball is positively charged, the electric force on it also points to the right. The tension in the thread makes the angle θ (= 30 ) with the vertical. ..................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... . . . . . . . . ~ T q ~ E m~g
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Unformatted text preview: yields qE − T sin θ = 0 and the sum of the vertical components yields T cos θ − mg = 0. The expression T = qE/ sin θ , from the Frst equation, is substituted into the second to obtain qE = mg tan θ . The electric Feld produced by a large uniform plane of charge is given by E = σ/ 2 ε , where σ is the surface charge density. Thus, qσ 2 ε = mg tan θ and σ = 2 ε mg tan θ q = 2(8 . 85 × 10 − 12 C 2 / N · m 2 )(1 . × 10 − 6 kg)(9 . 8 m / s 2 ) tan 30 ◦ 2 . × 10 − 8 C = 5 . × 10 − 9 C / m 2 ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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