Electric potential at the centre of a cube having uniform volumetric charge density ‘ $ρ$ ’ and side ‘a’ is given by V₀ . Then, electric potential at the tip of a square pyramid having same volumetric charge density ( $ρ$ ) of base area 4a² and height ‘a’ is:

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$\frac{V_{0}}{4}$

Can’t be determined

$\frac{V_{0}}{6}$

$\frac{2 V_{0}}{3}$

answer is C.

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Detailed Solution

$V_{0} = {kρa}^{2} =$ potential at the centre of a cube of side ‘a’
$kρ \times 4 a^{2} = 4 V_{0} =$ Potential at the centre of a cube of side ‘2a’
Six such pyramids make a cube
Potential at the tip of pyramid $\frac{4 V_{0}}{6} = \frac{2 V_{0}}{3}$ .