Question

Four point charges are placed on vertices of a

square of side 1 m as shown in the figure. The

net electric field at the centre of square is

(consider SI unit)
expalin it with solution...​

Answer 1

Given:

Four point charges are placed on the vertices of a square of side 1 metre.

To find:

Net Electrostatic Field at the centre of square.

Concept:

First refer to the attached diagram .

+q charges at point A and C shall produce equal and opposite field intensity. These will nullify each other and hence will not affect the net field intensity at the centre of the square.

Only -q charge at D and +q charge at B will produce equal field intensity in the same direction and hence the vectors will add up.

Calculation:

$\therefore \: E_{net} = E_{B} + E_{D}$

$= > \: E_{net} = \dfrac{1}{4\pi\epsilon_{0}} \dfrac{q}{ { (\frac{1}{ \sqrt{2} }) }^{2} } + \dfrac{1}{4\pi\epsilon_{0}} \dfrac{q}{ { (\frac{1}{ \sqrt{2} }) }^{2} }$

$= > \: E_{net} = \dfrac{1}{4\pi\epsilon_{0}} \dfrac{q}{ (\frac{1}{2} )} + \dfrac{1}{4\pi\epsilon_{0}} \dfrac{q}{( \frac{1}{2} )}$

$= > \: E_{net} = \dfrac{2q}{4\pi\epsilon_{0}} + \dfrac{2q}{4\pi\epsilon_{0}}$

$= > \: E_{net} = \dfrac{4q}{4\pi\epsilon_{0}}$

$= > \: E_{net} = \dfrac{q}{\pi\epsilon_{0}}$

So final answer :

Field intensity at centre of square will be:

$\boxed{ \sf{ \: E_{net} = \dfrac{q}{\pi\epsilon_{0}} }}$