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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsPotential & potential energyMedium2 minQB

PhysicsMediumsingle choice

Consider the potential at the corner of a uniformly charged cube of dimension L to be directly proportional to $ρ L^{2}$ , where $ρ$ is volume charge density. What is the value of the ratio of the potential at the centre to the potential at the corner of the cube :-

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Answer:

Solution:

$V_{A} \propto ρ L^{2}$ Divide cube into smaller cube of length $\frac{L}{2}$ Therefore

$V_{C} \propto 8 \cdot (ρ) (\frac{L}{2})^{2} \propto 2 ρ L^{2} V_{C} \propto 2 ρ L^{2} \frac{V _{A}}{V _{C}} = \frac{1}{2}$

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Potential & potential energy

⏱ 2mℹ️ Source: QB

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