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
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
- [JEE MAIN 2021]
- A
$\frac{-Q}{3 \sqrt{3} \pi \varepsilon_{0} a^{2}}(\hat{x}+\hat{y}+\hat{z})$
- B
$\frac{-2 Q}{3 \sqrt{3} \pi \varepsilon_{0} a^{2}}(\hat{x}+\hat{y}+\hat{z})$
- C
$\frac{2 Q}{3 \sqrt{3} \pi \varepsilon_{0} a^{2}}(\hat{x}+\hat{y}+\hat{z})$
- D
$\frac{ Q }{3 \sqrt{3} \pi \varepsilon_{0} a ^{2}}(\hat{ x }+\hat{ y }+\hat{ z })$
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- [JEE MAIN 2019]
Two point charges of $20\,\mu \,C$ and $80\,\mu \,C$ are $10\,cm$ apart. Where will the electric field strength be zero on the line joining the charges from $20\,\mu \,C$ charge......$m$
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$(a)$ Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where $E =0$ ) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
$(b)$ Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
$(a)$ Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where $E =0$ ) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
$(b)$ Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.