Question

charged gold for charged metal plates I Figure 9. Investigating the electric field between two charged metal plates 1 Figure
2 A charged dust particle in an electric field experiences a force of 44x10 N. The charge on the particle is 8.8x10-C. Calcu
13 3 Calculate the potential difference that must be applied across a pair of parallel plates, placed 4 cm apart, to produce
4 A potential difference of 2.4 KV is applied across a pair of parallel plates. The electric field strength berween the plate
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Answer #1

Charged dust Particle -13 N -17 c 4,4*10-15 Question 2 cho id F =4.4 * 10 q 8.8* 10 Thus, electric field strength E = F 4.4*1

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Question 3 - E » AV Potential difference Plate separation AV AR EAR - 4000 * 0.04 (.4 cm cm = 0.04 m) = 160 v = AV Answer c P

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