Question

three equal charges +q each are placed at the vertices of a square as shown in the figure. the intensity of electronic field at centre O will ​

Answer 1

Given:

Three equal charges +q each are placed at the vertices of a square as shown in the figure.

To find:

Field intensity at point O ?

Calculation:

• First see the hand-drawn diagram.

The field intensity due to the diagonally opposite charges (i.e. E1 and E2) are equal and opposite to one another.

So, E1 and E2 cancel out each other.

Net field intensity will be simply due to the third charge as follows:

$\therefore \: E_{net} = E_{3}$

$\implies \: E_{net} = \dfrac{q}{4\pi \epsilon_{0} {d}^{2} }$

• Here 'd' = ½ (diagonal) = a/√2.

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

$\implies \: E_{net} = \dfrac{2q}{4\pi \epsilon_{0} {a}^{2} }$

$\implies \: E_{net} = \dfrac{q}{2\pi \epsilon_{0} {a}^{2} }$

So, Option B) is correct !

Answer 2

Answer:

hope it helps... thanks plz