A particle having a charge of 15 µC is brought from infinity to a point P by doing a work of 3.0 mJ. The potential at point P is ________.

  1. 100 V
  2. 250 V
  3. 200 V
  4. 300 V

Answer (Detailed Solution Below)

Option 3 : 200 V
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Detailed Solution

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CONCEPT:

Electric potential (V):

  • The potential difference between two points in an electric field may be defined as the amount of work done in moving a unit positive charge from one point to the other against the electrostatic force i.e.,

\({\rm{Electric\;potential\;}}\left( {\rm{V}} \right) = \frac{{{\rm{Work\;done\;}}\left( {\rm{W}} \right)}}{{{\rm{Charge\;}}\left( {\rm{q}} \right)}}\)

  • The electric potential V at a point P due to point charge +q at a distance r is given by

\(⇒ V=\frac{1}{{4{\rm{\Pi }}\varepsilon_o }}\frac{q}{r}\)     

CALCULATION:

Given - Charge (q) = 15 μC = 15 × 10-6 C and work done (W) = 3 mJ = 3 × 10-3 J

  • The potential at point P is 

\(⇒ V =\frac{W}{q}=\frac{3\times 10^{-3}}{15\times 10^{-6}}=0.2\times 10^{3}\, V=200\, V\)

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