The work done in carrying a charge of $5 μ C$ from a point A to a point B in an electric field is 10 mJ. The potential difference $(V_{B} - V_{A})$ is

+ 2kV

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- 2 kV

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+ 200 V

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- 200 V

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Solution

The correct option is A + 2kV
Given,
Charge = 5 micro coulomb
Work done = 10 mJ
Now,
When a charge is carried from one point to another point then work done is given as
Work done $W = Q (V_{B} - V_{A}) \Rightarrow (V_{B} - V_{A}) = \frac{W}{Q} = \frac{10 \times 10^{- 3}}{5 \times 10^{- 6}} J / C = 2 k V$
Hence the potential difference is 2 kV

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The work done in carrying a charge of 5μC from a point A to a point B in an electric field is 10 mJ. The potential difference (VB−VA) is

The correct option is A + 2kVGiven,Charge = 5 micro coulombWork done = 10 mJNow,When a charge is carried from one point to another point then work done is given as Work done W=Q(VB−VA)⇒(VB−VA)=WQ=10×10−35×10−6J/C=2kVHence the potential difference is 2 kV

The work done in carrying a charge of $5 μ C$ from a point A to a point B in an electric field is 10 mJ. The potential difference $(V_{B} - V_{A})$ is