Q

# Confused! kindly explain, - Electrostatics - JEE Main-7

The distance between the two charges $25\mu C$  and  $36\mu C$  is 11 cm  At what point on the line joining the two, the intensity will be zero

• Option 1)

At a distance of 5 cm from $25\mu C$

• Option 2)

At a distance of 5 cm from $36\mu C$

• Option 3)

At a distance of 10 cm from $25\mu C$

• Option 4)

At a distance of 11 cm from $36\mu C$

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As we learned

Electric Field Intensity -

$\dpi{100} \vec{E}=\frac{\vec{F}}{q_{0}}=\frac{kQ}{r^{2}}$

- wherein

Suppose electric field is zero at point N in the figure then

At N |E1| = |E2 | which gives $x_{1}=\frac{x}{\sqrt{\frac{Q_2}{Q_1}}+1}=\frac{11}{\sqrt{\frac{36}{25}}+1}=5\; cm$

Option 1)

At a distance of 5 cm from $25\mu C$

Option 2)

At a distance of 5 cm from $36\mu C$

Option 3)

At a distance of 10 cm from $25\mu C$

Option 4)

At a distance of 11 cm from $36\mu C$

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