The potential at a point x (measured in μm ) due to some charges situated on the x-axis is given by V x = 20 x 2 - 4 volt. The electric field E at x = 4 μm is given by

Here, V x = 20 x 2 - 4 V As, E = - d V d x = - d d x 20 x 2 - 4 = + 20 1 x 2 - 4 2 ⋅ 2 x E = 40 x x 2 - 4 2 at x = 4 m , E = 40 × 4 4 2 - 4 2 = 160 144 = 10 9 V / μm in positive x -direction

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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsPotential & potential energyEasy2 minPYQ_2020

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The potential at a point $x$ (measured in $μm$ ) due to some charges situated on the x-axis is given by $V (x) = \frac{20}{(x^{2} - 4)}$ volt. The electric field $E$ at $x = 4 μm$ is given by

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Answer:

Solution:

Here, $V (x) = \frac{20}{x^{2} - 4} V$

As, $E = - \frac{d V}{d x}$

$= - \frac{d}{d x} (\frac{20}{x^{2} - 4})$

$= + 20 \{\frac{1}{{(x^{2} - 4)}^{2}}\} \cdot 2 x$

$E = \frac{40 x}{{(x^{2} - 4)}^{2}}$

at $x = 4 m,$

$E = \frac{40 \times 4}{{(4^{2} - 4)}^{2}}$

$= \frac{160}{144}$

$= \frac{10}{9} V / μm$ in positive $x$ -direction

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

⏱ 2mℹ️ Source: PYQ_2020

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