(a) A ball is attached to a thread of length L = 24.2 cm and suspended from the ceiling, as shown in the figure. A uniform electric field points to the right in the figure. When Ⓒ = 12.7°, the ball is in equilibrium. Find the net charge on the ball (in μC). A E=1.00 x 10³ N/C L m= 2.00 g Ⓡ 4.71 x Draw a free-body diagram, and determine the forces acting on the ball. Apply Newton's second law in the x- and y-directions, and solve your resulting equations for q. μC (b) What If? If the electric field is suddenly turned off, what is the speed of the ball at the bottom of its swing (in m/s)? 1x 0.375 Use energy conservation and solve your resulting equation for the speed v. m/s

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(a) A ball is attached to a thread of length L = 24.2 cm and suspended from the ceiling, as shown in the figure. A uniform electric field points to the right in the figure. When 8 = 12.7°, the ball is in equilibrium. Find the net charge on the ball (in µC).
m=
E = 1.00 × 10³ N/C
X
L
2.00 g
4.71
X
Draw a free-body diagram, and determine the forces acting on the ball. Apply Newton's second law in the x- and y-directions, and solve your resulting equations for q. µC
(b) What If? If the electric field is suddenly turned off, what is the speed of the ball at the bottom of its swing (in m/s)?
0.375
Use energy conservation and solve your resulting equation for the speed v. m/s
Transcribed Image Text:(a) A ball is attached to a thread of length L = 24.2 cm and suspended from the ceiling, as shown in the figure. A uniform electric field points to the right in the figure. When 8 = 12.7°, the ball is in equilibrium. Find the net charge on the ball (in µC). m= E = 1.00 × 10³ N/C X L 2.00 g 4.71 X Draw a free-body diagram, and determine the forces acting on the ball. Apply Newton's second law in the x- and y-directions, and solve your resulting equations for q. µC (b) What If? If the electric field is suddenly turned off, what is the speed of the ball at the bottom of its swing (in m/s)? 0.375 Use energy conservation and solve your resulting equation for the speed v. m/s
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