Question

find the magnitude and direction of electric field the p in fig​

Answer 1

Answer:

$2kq \div {a}^{2}$

Explanation:

The net electric field is in the direction of Electric field due to B

Answer 2

Concept:

The magnitude of the electric field (E) created at a point r distances distant from a point charge with a charge of magnitude Q.

Given:

There are three equal charges of $+q$ at points $A,B,C$ with length $AB$ and $BC$ as $a$ each.

Find:

The magnitude and direction of the electric field.

Solution:

The magnitude of the electric field is given as:

$E=\frac{kQ}{r^2}$

The charge, $Q=+q$

Since BP is perpendicular to AC,

$\Delta BCP$ is a right-angled triangle.

Now angle C is $45^o$.

Therefore, $cos 45^o=\frac{BP}{x}$

$\frac{1}{\sqrt{2} }=\frac{BP}{a}$

$BP=\frac{a}{\sqrt{2} }$

So,

$E=\frac{kq}{{\frac{a}{\sqrt{2} }}^2 }$

$E=\frac{k2q}{a^2}$

$E=\frac{1}{4\pi \varepsilon _0}\frac{2q}{a^2}$

Therefore, the magnitude of the electric field is $E=\frac{1}{4\pi \varepsilon _0}\frac{2q}{a^2}$  and the direction is along $BP$.

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