Two point charges 20mC and - 60 mC are separated by a distance of 16cm. The net electric potential is zero on the line joining them Options: A . &nbspAt a distance of 4cm from 20mC in between the charges B . &nbspAt a distance of 8cm from 20mC and 24cm from -60 mC C . &nbspBoth of the above D . &nbspNone of the above

Answer: Option C : C x = d q 2 q 1 ± 1 ; x 1 = d q 2 q 1 + 1 = 6 60 20 + 1 = 4 c m x 2 = 16 60 20 − 1 = 16 2 = 8 c m point outside

Two point charges 20mC and - 60 mC are separated by a distance of 16cm. Th

Question

Two point charges 20mC and - 60 mC are separated by a distance of 16cm. The net electric potential is zero on the line joining them

Options:

A . At a distance of 4cm from 20mC in between the charges

B . At a distance of 8cm from 20mC and 24cm from -60 mC

C . Both of the above

D . None of the above

Answer: Option C
:
C

x = \frac{d}{\frac{q_{2}}{q_{1}} \pm 1}; x_{1} = \frac{d}{\frac{q_{2}}{q_{1}} + 1} = \frac{6}{\frac{60}{20} + 1} = 4 c m x_{2} = \frac{16}{\frac{60}{20} - 1} = \frac{16}{2} = 8 c m

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