An electrical charge of 2μC is placed at the point (1,2,3). At the point (2,3,4) the electric field and potential will be :
6×103NC−1 and 6×103JC−1
6000NC−1 and 6000√3JC−1
6×103NC−1 and 3√3JC−1
none of the above
A
6000NC−1 and 6000√3JC−1
B
6×103NC−1 and 6×103JC−1
C
none of the above
D
6×103NC−1 and 3√3JC−1
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Solution
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The electric field ,
E=14πϵ0qr2=9×109×2×10−6(2−1)2+(3−2)2+(4−3)2=6000NC−1
and potential,
V=14πϵ0qr=9×109×2×10−6√(2−1)2+(3−2)2+(4−3)2=6000√3NC−1
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