At any point x , 0 , 0 the electric potential V is 1000 x + 1500 x 2 + 500 x 3 volt , then electric field intensity at x = 1 m is

Electric potential V = 1000 x + 1500 x 2 + 500 x 3 Electric field E = - dV dx = 1000 x 2 + 2 1500 x 3 + 3 500 x 4 i ^ At x = 1 , E = 5500 i ^ V m - 1

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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsPotential & potential energyEasy2 minPYQ_2020

PhysicsEasysingle choice

At any point $(x, 0, 0)$ the electric potential $V$ is $(\frac{1000}{x} + \frac{1500}{x^{2}} + \frac{500}{x^{3}}) volt$ , then electric field intensity at $x = 1 m$ is

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Answer:

Solution:

Electric potential

$V = (\frac{1000}{x} + \frac{1500}{x^{2}} + \frac{500}{x^{3}})$

Electric field

$E = - \frac{dV}{dx} = (\frac{1000}{x^{2}} + \frac{(2) 1500}{x^{3}} + \frac{(3) 500}{x^{4}}) \hat{i}$

At $x = 1$ ,

$E = 5500 \hat{i} V m^{- 1}$

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

⏱ 2mℹ️ Source: PYQ_2020

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