Electric potential at a point 'P' due to a point charge of 5 × 1 0 − 9 C is 50 V . The distance of ' P ' from the point charge is: (Assume, 4 π ε 0 1 = 9 × 1 0 + 9 Nm 2 C − 2 )

Electric potential at a point P due to a point charge, ( ∵ K = 9 × 1 0 9 ) V P = r K Q ⇒ 50 = r 9 × 1 0 9 × 5 × 1 0 − 9 ⇒ r = 50 45 = 10 9 = 0.9 m = 90 cm

Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsPotential & potential energyMedium2 minPYQ_2024

PhysicsMediumsingle choice

Electric potential at a point 'P' due to a point charge of $5 \times 1 0^{- 9} C$ is 50 V . The distance of ' P ' from the point charge is: (Assume, $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{+ 9} Nm^{2} C^{- 2}$ )

Options:

Answer:

Solution:

Electric potential at a point P due to a point charge, $(∵ K = 9 \times 1 0^{9})$ $V_{P} = \frac{K Q}{r}$ $\Rightarrow 50 = \frac{9 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 9}}{r}$ $\Rightarrow r = \frac{45}{50} = \frac{9}{10} = 0.9 m = 90 cm$

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

⏱ 2mℹ️ Source: PYQ_2024

Doubts & Discussion

Loading discussions...