Electrostatic Potential and Capacitance · Class 12 · CBSE · Science Physics

The electric potential at any point (x, y, z) in space is given by V = 3x^2 + y, where V is in volts and x, y are in meters. The magnitude of the ele…

Practice question from Electrostatic Potential and Capacitance (Class 12 · CBSE · Science Physics), with the correct answer and a step-by-step solution.

medium · MCQ · Q-54D74E

The electric potential at any point (x,y,z)(x, y, z) in space is given by V=3x2+yV = 3x^2 + y, where VV is in volts and x,yx, y are in meters. The magnitude of the electric field at the point (1 m,2 m,0 m)(1\text{ m}, 2\text{ m}, 0\text{ m}) is:

  1. A.6 V/m6\text{ V/m}
  2. B.7 V/m7\text{ V/m}
  3. C.37 V/m\sqrt{37}\text{ V/m}Correct answer
  4. D.3 V/m3\text{ V/m}

Solution

The components of the electric field are given by Ex=−∂V∂x=−6xE_x = -\frac{\partial V}{\partial x} = -6x, Ey=−∂V∂y=−1E_y = -\frac{\partial V}{\partial y} = -1, and Ez=0E_z = 0. At point (1,2,0)(1, 2, 0), Ex=−6(1)=−6 V/mE_x = -6(1) = -6\text{ V/m} and Ey=−1 V/mE_y = -1\text{ V/m}. The magnitude of the electric field is E=Ex2+Ey2+Ez2=(−6)2+(−1)2+0=36+1=37 V/mE = \sqrt{E_x^2 + E_y^2 + E_z^2} = \sqrt{(-6)^2 + (-1)^2 + 0} = \sqrt{36 + 1} = \sqrt{37}\text{ V/m}.

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