2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Electrostatics & Capacitance

JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Self Evaluation Test - Electrostatic Potential and Capacitance

  • question_answer
    The potential at the point x (measured in\[\mu m\]) due to some charges situated on the x-axis is given by \[V(x)=20({{x}^{2}}-4)\] volt. The electric field E at \[x=4\mu m\]

    A) \[\left( 10/9 \right)\text{ }volt/\mu m\]and in the +ve x direction

    B) \[\left( 5/3 \right)\text{ }volt/\mu m\]and in the -ve x direction

    C) \[\left( 5/3 \right)\text{ }volt/\mu m\] and in the +ve x direction

    D) \[\left( 10/9 \right)\text{ }volt/\mu m\]and in the -ve x direction

    Correct Answer: A

    Solution :

    [a] Here, \[V\left( x \right)=\frac{20}{{{x}^{2}}-4}volt\] We know that  \[E=-\frac{dV}{dx}=-\frac{d}{dx}\left( \frac{20}{{{x}^{2}}-4} \right)\] \[\text{or, }E=+\frac{40x}{{{\left( {{x}^{2}}-4 \right)}^{2}}}\] At\[x=4\mu m\], \[E=+\frac{40\times 4}{{{({{4}^{2}}-4)}^{2}}}=+\frac{160}{144}=+\frac{10}{9}volt/\mu m\]. Positive sign indicates that \[\vec{E}\] is in +ve x-direction.


You need to login to perform this action.
You will be redirected in 3 sec spinner