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The work done in carrying a charge of 5 μC from point A to a point B in an electric field is 10mJ. The potential difference VB- VA is then
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a
+2 KV.
b
-2 KV
c
+ 200 V
d
-200 V
answer is B.
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Detailed Solution
wAB=5μCVA−VB⇒10mJ5μC=VA−VB ∴ VA−VB=2kV or VB−VA=−2kV
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