Two identical charges of value Q each are placed at (.-a, 0.) and (.a, 0.) . The coordinates of the points where the net electric field is zero and maximum are respectively -

Question

Two identical charges of value Q each are placed at (.-a, 0.) and (.a, 0.) . The coordinates of the points where the net electric field is zero and maximum are respectively - (A) (0 , 0), (0 , 0) (B) (0 , (a/√2)), (0 , 0)  (C) (0 , 0), (0 , (a/√2))  (D) ((a/√2) , 0), (0 , (a/√2))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-0faa4hkj0mub.png" /> Net electric field will be zero at origin. At any co-ordinate (0 , y) EN e t=2E(c o s 2 θ /2) =(k Q y/(a2 + y2)(3)2) For maximum electric field (d E/d y)=0 Solving y=±(a/√2)

Q. Two identical charges of value $Q$ each are placed at $(- a, 0)$ and $(a, 0)$ . The coordinates of the points where the net electric field is zero and maximum are respectively -

$(0, 0), (0, 0)$

$(0, \frac{a}{2}), (0, 0)$

$(0, 0), (0, \frac{a}{2})$

$(\frac{a}{2}, 0), (0, \frac{a}{2})$

Net electric field will be zero at origin.
At any co-ordinate $(0, y)$ $E_{N e t} = 2 E \frac{cos 2 θ}{2}$
$= \frac{k Q y}{( a ^{2} + y ^{2} ) ^{\frac{3}{2}}}$
For maximum electric field $\frac{d E}{d y} = 0$
Solving $y = \pm \frac{a}{2}$

Q. Two identical charges of value Q each are placed at (−a,0) and (a,0) . The coordinates of the points where the net electric field is zero and maximum are respectively -

(0,0),(0,0)

(0,2​a​),(0,0)

(0,0),(0,2​a​)

(2​a​,0),(0,2​a​)