Question

in the following figure the electric field on the y axis will be maximum at Y equal to ​

Answer 1

Given:

2 charges have been placed.

To find:

Value of y for which Field Intensity will be max?

Calculation:

Let net field intensity at P be E⁰ :

$\therefore \: {E}^{o} = 2 \times E \times \cos( \theta)$

$\implies\: {E}^{o} = 2 \times \dfrac{kq}{ {r}^{2} } \times \cos( \theta)$

$\implies\: {E}^{o} = 2 \times \dfrac{kq}{ {( \sqrt{ { d }^{2} + {y}^{2} } )}^{2} } \times \dfrac{y}{ \sqrt{ {d}^{2} + {y}^{2} } }$

$\implies\: {E}^{o} = \dfrac{2kqy}{ {({ d }^{2} + {y}^{2} )}^{ \frac{3}{2} } }$

Now, differentiation w.r.t to y and equating to zero for maxima :

$\implies\: \dfrac{d{E}^{o}}{dy} = 0$

$\implies\: {( {d}^{2} + {y}^{2} )}^{ \frac{3}{2} } - y \bigg\{ \dfrac{3}{2} {( {d}^{2} + {y}^{2} )}^{ \frac{1}{2} } \times 2y \bigg\} = 0$

$\implies\: {( {d}^{2} + {y}^{2} )}^{ \frac{3}{2} } = \bigg\{ 3 {y}^{2} {( {d}^{2} + {y}^{2} )}^{ \frac{1}{2} } \bigg\}$

$\implies\: {( {d}^{2} + {y}^{2} )}^{1} =3 {y}^{2}$

$\implies\: {d}^{2} + {y}^{2} =3 {y}^{2}$

$\implies\: 2{y}^{2} ={d}^{2}$

$\implies\: {y}^{2} = \dfrac{{d}^{2}}{2}$

$\implies\: y = \dfrac{d}{ \sqrt{2} }$

So, maxima Intensity if y = d/√2.

Answer 2

Explanation:

here is the answer hope it helps..