Question

A cube of side 'a' has point charges +Q located at each of its vertices except at the origin where the charge is –Q. The electric field at the centre of cube is :

Answer 1

Correct option is (2) \( \frac{-2Q}{3\sqrt 3 \pi \varepsilon_0a^2} (\hat x + \hat y + \hat z)\)

If we consider two point charges +Q and –2Q at origin,  then +Q charge along with other seven +Q charges will produce zero net electric field at centre of cube due to symmetry. Now only remaining charge is –2Q at center. 

So net electric field at centre will only be due to -2Q charge.

\(\vec E = \frac{kq}{r^3} \vec r\) 

Here, \(\vec r = \frac a2 \hat i + \frac a2 \hat j + \frac a2\hat k\)

⇒ \(\vec E = \frac{K - (2Q)}{\left(\frac{a\sqrt 3}2\right)^3} \left(\frac a2 \hat i + \frac a2\hat j + \frac a2\hat k\right)\)

⇒ \(\vec E = \frac{-2Q}{3\sqrt 3 \pi \varepsilon_0a^2} (\hat i + \hat j + \hat k)\)

We can replace unit vectors \(\hat i , \hat j, \hat k \) with \(\hat x, \hat y, \hat z\).

Answer 2

Correct option (2)

We can replace –Q charge at origin by +Q and –2Q . Now due to +Q charge at every corner of cube. Electric field at center of cube is zero so now net electric field at center is only due to –2Q charge at origin.