Question

Two identical charges of value Q each are placed at (-a,0) and (a,0). The co-ordinates of the points where 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)​

Answer 1

Answer:

I think the answer is c

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Answer 2

Given:

Two identical charges of value Q each are placed at (-a,0) and (a,0).

To find:

The co-ordinates of the points where net electric field is zero and maximum.

Calculation:

The net field intensity at a distance x from the origin (0,0) along X AXIS:

$\therefore \: E_{net} = \dfrac{kq}{ {(a - x)}^{2} } - \dfrac{kq}{ {(a + x)}^{2} }$

$= > \: E_{net} =kq \bigg \{ \dfrac{1}{ {(a - x)}^{2} } - \dfrac{1}{ {(a + x)}^{2} } \bigg \}$

$= > \: E_{net} =kq \bigg \{ \dfrac{ {(a + x)}^{2} - {(a - x)}^{2} }{ {( {a}^{2} - {x}^{2} )}^{2} } \bigg \}$

$= > \: E_{net} =kq \bigg \{ \dfrac{4ax }{ {( {a}^{2} - {x}^{2} )}^{2} } \bigg \}$

Now , we can see that min E_{net} will be at origin , when value of x = 0.

So, E_(net) = 0 at (0,0).

Electrostatic field intensity along the y axis :

$\therefore \: E_{net} = 2 \dfrac{kq}{ {( \sqrt{ {a}^{2} + {y}^{2} }) }^{2} } \cos( \theta)$

$= > \: E_{net} = 2 \dfrac{kq}{ {( \sqrt{ {a}^{2} + {y}^{2} }) }^{2} } \times \dfrac{y}{ \sqrt{ {a}^{2} + {x}^{2} } }$

$= > \: E_{net} = \dfrac{2kqy}{ { ({a}^{2} - {x}^{2} ) }^{ \frac{3}{2} } }$

In this expression , max value comes at y = a/√2.

So max E_(net) is at (0,a/√2).

HOPE IT HELPS.

• You do need to remember these values directly for competitive exams and the proof for subjective.