Q. 14.4( 42 Votes )
Two charges qA = 5 x 10-8 C and qB = -3x10-8 C
Distance between two charges, r = 16 cm = 0.16 m
Consider a point O on the line joining two charges where the electric potential is zero due to two charges.
From the figure we can see that, x = distance of point O from charge qA
Electric potential at point O due to qA,
ϵ0 is the permittivity of free space. Its value is 8.85 x 10-12 F/m
Electric potential at point O due to qB
Since the total electric potential at O is zero,
⇒ VA + VB = 0
On cross multiplying we get,
5 × (0.16 -x) = 3x
⇒ 0.8 – 5x = 3x
⇒ 8x = 0.8
⇒x = 0.1 m = 10 cm(from charge qA)
∴ at a distance of 10 cm from the positive charge, the potential is zero between the two charges.
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