Question

Please solve this problem.​

Answer 1

Electric field:

• region created by electric charges.
• it is a vector quantity.
• direction is away from positive charge and it is towards the negative charge.

Solution:

First, draw the figure according to question.

• At middle point EF due to positive charge will be +x axis direction.
• and due to negative charge also in +x axis.

$\boxed{\rm \vec{E} = \dfrac{Kq}{r^2} =\dfrac{q}{4\pi \epsilon_0 r^2}}$

$\rm q = 0.01\mu C , r = \dfrac{10}{2}= 5cm$

Electric field intensity at middle point:

$\longrightarrow \rm \vec{E}_{net} = \vec{E}_1 +\vec{E}_2$

$\longrightarrow \rm \vec{E}_{net} = \dfrac{Kq}{r^2} \hat{i} + \dfrac{Kq}{r^2} \hat{i}$

$\longrightarrow \rm \vec{E}_{net} = \dfrac{2Kq}{r^2} \hat{i}$

$\longrightarrow \rm \vec{E}_{net} = \dfrac{2\times 9\times 10^9 \times 0.01 \times 10^{-6}}{(5\times 10^{-2})^2} \hat{i}$

$\longrightarrow \rm \vec{E}_{net} = \dfrac{2\times 9\times 10}{25\times 10^{-4}} \hat{i}$

$\longrightarrow \rm \vec{E}_{net} = \dfrac{180\times 10^4}{25} \hat{i}$

$\implies \bf \vec{E}_{net} = 7.2 \times 10^4 \: \hat{i} \: \dfrac{N}{C}$

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☆ note:

• negative sign of charge is only used for direction not in magnitude.