JEE MAIN - Physics (2021 - 24th February Morning Shift - No. 1)
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 :
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$${{2Q} \over {3\sqrt 3 \pi {\varepsilon _0}{a^2}}}\left( {\widehat x + \widehat y + \widehat z} \right)$$
$${{ - Q} \over {3\sqrt 3 \pi {\varepsilon _0}{a^2}}}\left( {\widehat x + \widehat y + \widehat z} \right)$$
$${Q \over {3\sqrt 3 \pi {\varepsilon _0}{a^2}}}\left( {\widehat x + \widehat y + \widehat z} \right)$$
$${{ - 2Q} \over {3\sqrt 3 \pi {\varepsilon _0}{a^2}}}\left( {\widehat x + \widehat y + \widehat z} \right)$$
Explanation
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,
$$E = {{Kqr} \over {{r^3}}}$$
Here, position vector, $$r = {a \over 2}(\widehat x + \widehat y + \widehat z)$$
$$ \Rightarrow \left| r \right| = \sqrt {{{\left( {{a \over 2}} \right)}^2} + {{\left( {{a \over 2}} \right)}^2} + {{\left( {{a \over 2}} \right)}^2}} = {{\sqrt 3 a} \over 2}$$
$$ \Rightarrow E = {1 \over {4\pi {\varepsilon _0}}} \times {{( - 2Q)a/2} \over {{{\left( {{a \over 2}\sqrt 3 } \right)}^3}}}\,.\,(\widehat x + \widehat y + \widehat z)$$
$$E = {{-2Q(\widehat x + \widehat y + \widehat z)} \over {3\sqrt 3 \pi {a^2}{\varepsilon _0}}}$$
$$E = {{Kqr} \over {{r^3}}}$$
Here, position vector, $$r = {a \over 2}(\widehat x + \widehat y + \widehat z)$$
$$ \Rightarrow \left| r \right| = \sqrt {{{\left( {{a \over 2}} \right)}^2} + {{\left( {{a \over 2}} \right)}^2} + {{\left( {{a \over 2}} \right)}^2}} = {{\sqrt 3 a} \over 2}$$
$$ \Rightarrow E = {1 \over {4\pi {\varepsilon _0}}} \times {{( - 2Q)a/2} \over {{{\left( {{a \over 2}\sqrt 3 } \right)}^3}}}\,.\,(\widehat x + \widehat y + \widehat z)$$
$$E = {{-2Q(\widehat x + \widehat y + \widehat z)} \over {3\sqrt 3 \pi {a^2}{\varepsilon _0}}}$$
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