Answer :


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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