100J of work is done when 2μC charge is moved in an electric field between two points. The p.d. between the points is
2×10−4V
2×10−8V
2×10−6V
5×107V
A
2×10−8V
B
2×10−4V
C
2×10−6V
D
5×107V
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Solution
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We know work done =qΔV Where ΔV is change in potential V2−V1 100=ΔV×2×10−6 50×106=ΔV ∴p⋅d=5×107V
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