Question

two equal point charges A and B are R distance apart. A third point charge placed on the prependicular bisector at a distance 'd' from the center will experience maximum electrostatic force when

1. d = R/2 X 1.414

2 d = R/ 1.414

3. d = R 1.414

4. d = 2 X 1.414 R

Answer 1

If you want to do it fast Keep it very simple

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As we know,

Electrostatic Force is inversely proportional to radius

So the option which has Min radius will have Max electrostatic force

So, the option would be

R/2*1.414

If you go on solving it will waste minimum 5 on ur exam whether it may be NEET OR JEE

Some questions are purely based on concept even if the include number

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Answer 2

Answer:

The correct answer is option (1). $\bold{D=\frac{R}{2 \times 1.414}}$

Explanation:

The two point charges of equal magnitude are placed at a distance of R from each other, with introduction of third point charge, electrostatic force calculation are as follows -

Given that the point charges are A and B

Distance between A and B = R

As we know electrostatic force is given by –

$\bold{F=\frac{k q^{2}}{r^{2}}}$

With third charge at perpendicular distance, we have –

$\bold{F=\frac{2 \times k q^{2}}{r^{2} \times \sin \theta}}$

$\begin{array}{l}{\Rightarrow F=\frac{2 \times k q^{2}}{r^{2} \times \frac{d}{r}}} \\ {\Rightarrow F=\left(\frac{2 k q^{2}}{\left[d^{2}+\left(\frac{R}{2}\right)^{2}\right]^{\frac{3}{2}}}\right)}\end{array}$

$\Rightarrow F(d)=\frac{K d}{\left(\sqrt{d^{2}}+\frac{R^{2}}{4}\right)^{3}}$

Now to have maximum electrostatic force,

F’ ( d ) = 0

$\begin{array}{l}{\Rightarrow\left(\sqrt{d^{2}}+\frac{R^{2}}{4}\right)^{3} \times 1-d \times \frac{3}{2}\left(\sqrt{d^{2}}+\frac{R^{2}}{4}\right)^{\frac{1}{2}} \times 2 d=0} \\ {\Rightarrow\left(\sqrt{d^{2}}+\frac{R^{2}}{4}\right)-3 d^{2}=0} \\ {\Rightarrow D=\frac{R}{2 \sqrt{2}}} \\ {\Rightarrow \bold{D=\frac{R}{2 \times 1.414}}}\end{array}$