Two point charges 20 mC and - 60 mC are separated by a distance of 16 cm- The net electric potential is zero on the line joining themA- At a distance of 4 cm from 20 mC in between the chargesB- At a distance of 8 cm from 20 mC and 24 cm from -60 mCC- Both of the aboveD- None of the above

The correct option is C Both of the abovex=d/q_2/q_1± 1;x_1=d/q_2/q_1+1=6/60/20+1=4cm x_2=16/60/20 1=16/2=8cm point outside ...

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Standard XII

Physics

Electric Potential at a Point in Space Due to a Point Charge Q

Two point cha...

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

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

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

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Both of the above

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None of the above

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Solution

The correct option is C Both of the above
$x = \frac{d}{\frac{q_{2}}{q_{1}} \pm 1}; x_{1} = \frac{d}{\frac{q_{2}}{q_{1}} + 1} = \frac{6}{\frac{60}{20} + 1} = 4 c m x_{2} = \frac{16}{\frac{60}{20} - 1} = \frac{16}{2} = 8 c m$
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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

The correct option is C Both of the abovex=dq2q1±1;x1=dq2q1+1=66020+1=4cmx2=166020−1=162=8cm point outside

The correct option is C Both of the above
$x = \frac{d}{\frac{q_{2}}{q_{1}} \pm 1}; x_{1} = \frac{d}{\frac{q_{2}}{q_{1}} + 1} = \frac{6}{\frac{60}{20} + 1} = 4 c m x_{2} = \frac{16}{\frac{60}{20} - 1} = \frac{16}{2} = 8 c m$
point outside