9. -1 points SerPSE9 25.P.005.WI. My Notes A uniform electric field of magnitude
ID: 1881414 • Letter: 9
Question
9. -1 points SerPSE9 25.P.005.WI. My Notes A uniform electric field of magnitude 345 V/m is directed in the negative y direction as shown in the figure below. The coordinates of point are (-0.500,-0.350) m, and those of point () are (0.700, 0.900) m. Calculate the electric potential difference VB-VA using the dashed-line path. Need Help?eWacht 10. 1 SerPSE9 25.P.009. Notes A particle having charge q = +1.90 C and mass m = 0.010 0 kg is connected to a string that is L = 1.10 m long and tied to the pivot point Pin the figure below. The particle, string, and pivot point all lie on a frictionless, horizontal table. The particle is released from rest when the string makes an angle = 60.0° with a uniform electric field of magnitude E = 310 V/m. Determine the speed of the particle when the string is parallel to the electric field m/s Top view Need Help?R 11. -2 points SerPSE9 25.P.052.MI My Notes Lightning can be studied with a Van de Graaff generator, which consists of a spherical dome on which charge is continuously deposited by a moving belt. Charge can be added until the electric field at the surface of the dome becomes equal to the dielectric strength of air. Any more charge leaks off in sparks as shown in the figure below. Assume the dome has a diameter of 46.0 cm and is surrounded by dry air with a "breakdown" electric field of 3.00 x 106 V/m (a) What is the maximum potential of the dome? kv b) What is the maximum charge on the dome? AC Need Help? imeniExplanation / Answer
V= Ed
= 345*( 0.9-(-0.35))
= 431.25 V
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10.
Using conservation of energy
Electric potential energy= kinetic energy gained
qV= 0.5mv^2
q(Ed)= mv^2
qE* L cos x= 0.5mv^2
1.9*10^-6*310*1.1* cos 60= 0.5*0.01* v^2
v= 0.2545 m/s
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Plz post other problems seperately
Comment in case any doubt.. good luck
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