Question

please solve..........​

Answer 1

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Due to infinitely long plane carrying surface charge density σσ, electric field at any point is given by

E=σ2∈0E=σ2∈0

Due to sheet 1,

E1=3σ2∈0E1=3σ2∈0(−j^)(−j^)

Due to sheet 2,

E2=2σ2∈0E2=2σ2∈0(−j^)(−j^)

Due to sheet 3,

E3=−σ2∈0E3=−σ2∈0(−j^)(−j^)

E1+E2+E3=−6σ2∈0j^E1+E2+E3=−6σ2∈0j^

=>−3σ∈0−3σ∈0j^j^

Hence B is the correct answer

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Answer 2

Answer:

Given:

Three infinitely long charged sheets are placed as shown in the figure.

To find:

Field intensity at point P according to the co-ordinate axes.

Calculation:

The net electrostatic field intensity at point P will be superimposition of the field vectors from the individual charged sheets at point P.

The upper-most plate has positive surface charge density and hence the field intensity of that plate at point P will be directed towards -Z axis.

The middle and lower plate has negative surface charge density hence the field intensity for those plates at point P will also be directed towards -Z axis.

Let net field be

$\sf{ \therefore\vec{E}_{net} = \dfrac{ \sigma}{ 2\epsilon_{0}} (- \hat{k}) + \dfrac{ 2\sigma}{ 2\epsilon_{0}} (- \hat{k}) + \dfrac{ \sigma}{ 2\epsilon_{0}} (- \hat{k})}$

$\sf{ = > \vec{E}_{net} = \dfrac{ 4\sigma}{ 2\epsilon_{0}} (- \hat{k}) }$

$\sf{ = > \vec{E}_{net} = \dfrac{ 2\sigma}{ \epsilon_{0}} (- \hat{k}) }</p><p>$

So, final answer is:

$\boxed{ \red{ \large{\rm{\vec{E}_{net} = \dfrac{ 2\sigma}{ \epsilon_{0}} (- \hat{k}) }}}} </p><p>$