Joint Entrance Examination/AIEEE 2009/Q064 POTENTIAL

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Original Question by 164.100.170.4 Here's the question... Q 64.Two point P and Q are maintained at the potentials of 10 V and –4 V respectively. The work done in moving 100 electrons from P to Q is (1) 9.60 × 10–17 J (2) –2.24 × 10–16 J (3) 2.24 × 10–16 J (4) – 9.60 × 10-17 J Posted by 164.100.170.4 04:52, 24 September 2009 (UTC) Post a reply![1]

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Reply to the thread, posted by 164.100.170.4 Thread post Additional notes Solution : Q = 100e = –100 × 1.6 × 10–19 = –1.6 × 10–17C ΔV = –14 V ∴ W = QΔ V = 14 × 1.6 × 10–17 = 2.24 × 10–16 J These comments, along with the notes on the left, were contributed by 164.100.170.4 04:53, 24 September 2009 (UTC)

• reply[2]

Reply to the thread, posted by 164.100.170.4 Thread post Additional notes Solution : Q = 100e = –100 × 1.6 × 10–19 = –1.6 × 10–17C ΔV = –14 V ∴ W = QΔ V = 14 × 1.6 × 10–17 = 2.24 × 10–16 J These comments, along with the notes on the left, were contributed by 164.100.170.4 04:55, 24 September 2009 (UTC)

• reply[3]

[[Category:AIEEE 2009]

Reply to the thread, posted by 117.199.161.166 Thread post the potential differenc between the two points is 14V . work done is q*V.Q=-100*1.6*10-19.W=-100*1.6*10-19*14 =-2.24 * 10-16 Additional notes These comments, along with the notes on the left, were contributed by 117.199.161.166 19:58, 16 December 2009 (UTC)

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