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  • Can someone explain Two point charges – Q and 2Q are separated by a distance R, neutral point will be obtained at

Two point charges – Q and 2Q are separated by a distance R, neutral point will be obtained at 
 

  • Option 1)

    A distance of \frac{R}{\sqrt{2-1}}  from – Q charge and lies between the charges.

  • Option 2)

    A distance of \frac{R}{\sqrt{2-1}} from – Q charge on the left side of it 
     

  • Option 3)

    A distance of \frac{R}{\sqrt{2-1}} from 2Q charge on the right side of it
     

  • Option 4)

    A point on the line which passes perpendicularly through the centre of the line joining – Q and 2Q charge.
     

 

Answers (2)

best_answer

As we learned

 

Zero potential due to a system of two point charge -

If both charges are like then resultant potential is not zero at any finite point.

If the charges are unequal and unlike then all such points where resultant potential is zero lies on a closed curve

- wherein

For internal point

 \left ( It \: is \: assumed\: that\left | Q_{1} \right |< \left | Q_{2} \right | \right )

At \\*P,\: \: \frac{Q_{1}}{x_{1}}= \frac{Q_{2}}{\left ( x-x_{1} \right )} \\* \Rightarrow x_{1}= \frac{x}{\left ( Q_{2}/Q_{1}+1 \right )}

 

 As already we discussed neutral point will be obtained on the side of charge which is smaller in magnitude i.e. it will obtained on the left side of – Q charge and at a distance.

t=\frac{R}{\sqrt{\frac{2Q}{Q}}-1}\Rightarrow t=\frac{R}{\sqrt{2}-1}


Option 1)

A distance of \frac{R}{\sqrt{2-1}}  from – Q charge and lies between the charges.

Option 2)

A distance of \frac{R}{\sqrt{2-1}} from – Q charge on the left side of it 
 

Option 3)

A distance of \frac{R}{\sqrt{2-1}} from 2Q charge on the right side of it
 

Option 4)

A point on the line which passes perpendicularly through the centre of the line joining – Q and 2Q charge.
 

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