Within a spherical charge distribution of charge density ρ r , N equipotential surfaces of potential V 0 , V 0 + ∆ V , V 0 + 2 ∆ V , … V 0 + N ∆ V ∆ V > 0 , are drawn and have increasing radii r 0 , r 1 , r 2 , … r N , respectively. If the difference in the radii of the surfaces is constant for all values of V 0 a n d ∆ V then :

E = − Δ V Δ r ∴ △v and △r are constant E = constant E = K q r 2 = c ⇒ q ∝ r 2 ∴ q = ∫ 0 r ρ 4 π r 2 d r ⇒ ρ ∝ 1 r

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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsPotential & potential energyHard2 minPYQ_2016

PhysicsHardsingle choice

Within a spherical charge distribution of charge density $ρ (r),$ N equipotential surfaces of potential $V_{0}, V_{0} + ∆ V, V_{0} + 2 ∆ V, \dots V_{0} + N ∆ V (∆ V > 0),$ are drawn and have increasing radii $r_{0}, r_{1}, r_{2}, \dots r_{N},$ respectively. If the difference in the radii of the surfaces is constant for all values of $V_{0} a n d ∆ V$ then :

Options:

Answer:

Solution:

$E = - \frac{Δ V}{Δ r}$

$∴$ △v and △r are constant

E = constant

$E = \frac{K q}{r^{2}} = c$

$\Rightarrow q \propto r^{2}$

$∴ q = \int_{0}^{r} ρ 4 π r^{2} d r$

$\Rightarrow ρ \propto \frac{1}{r}$

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

⏱ 2mℹ️ Source: PYQ_2016

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