Test Index

Electrostatics Part 1

Section: Physics

Question : 8 of 100

Marks: +1, -0

A cube of side

a

+Q

-Q

$\;\frac{-Q}{3\sqrt{3}\pi \varepsilon_0 a^2}\left(\hat{x}+\hat{y}+\hat{z}\right)$
$\;\frac{Q}{3\sqrt{3}\pi \varepsilon_0 a^2}\left(\hat{x}+\hat{y}+\hat{z}\right)$
$\;\frac{-2Q}{3\sqrt{3}\pi \varepsilon_0 a^2}\left(\hat{x}+\hat{y}+\hat{z}\right)$
$\;\frac{2Q}{3\sqrt{3}\pi \varepsilon_0 a^2}\left(\hat{x}+\hat{y}+\hat{z}\right)$

Solution:

We can replace

-Q

charge at origin by

+Q

and

-2Q.

Now, due to

+Q

charge at every corner of cube, electric field at centre of cube is zero.

So, net electric field at centre is only due to

-2Q

charge at origin. Vector form of electric field strength,

E=\;\frac{K q r}{r^{3}}

Here, position vector,

r=\;\frac{a}{2}\left(\hat{x}+\hat{y}+\hat{z}\right)

\Rightarrow \;\; |r| = \sqrt{\left(\;\frac{a}{2}\right)^2 + \left(\;\frac{a}{2}\right)^2 + \left(\;\frac{a}{2}\right)^2} = \frac{\sqrt{3} a}{2}

\Rightarrow \;\; E = \frac{1}{4\pi\varepsilon_0} \times \frac{ \frac{(-2Q)a}{2} }{ \left( \frac{a}{2} \sqrt{3} \right)^3 } \cdot \left( \hat{x} + \hat{y} + \hat{z} \right)

E = \frac{-2Q\left(\hat{x}+\hat{y}+\hat{z}\right)}{3\sqrt{3}\pi a^2 \varepsilon_0}

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