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:

Option 1 -

$\frac{2 Q}{3 \sqrt[]{3} π ε_{0} a^{2}} (\hat{x} + \hat{y} + \hat{z})$

Option 2 -

$\frac{Q}{3 \sqrt[]{3} π ε_{0} a^{2}} (\hat{x} + \hat{y} + \hat{z})$

Option 3 -

$\frac{- 2 Q}{3 \sqrt[]{3} π ε_{0} a^{2}} (\hat{x} + \hat{y} + \hat{z})$

Option 4 -

$\frac{- Q}{3 \sqrt[]{3} π ε_{0} a^{2}} (\hat{x} + \hat{y} + \hat{z})$

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Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

If charge (-Q) at origin is replaced by (+Q), then electric field at the centre of the cube is zero. Thus, electric field at the centre of the cube is as if only (-2Q) charge is present at the origin.

$\vec{E} = \frac{1}{4 π ε_{0}} \frac{(- 2 Q)}{{(\frac{\sqrt[]{3}}{2} a)}^{2}} . \frac{1}{\sqrt[]{3}} (\hat{x} + \hat{y} + \hat{z}) = \frac{| - 2 Q}{3 \sqrt[]{3} π ε_{0} a^{2}} (\hat{x} + \hat{y} + \hat{z})$

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