Electrostatics — Medium JEE physics MCQ
The electric potential in a region is given by \(V = 3x^2 + 4y\) volts, where \(x\) and \(y\) are in meters. The magnitude of the electric field at the point \((1, 2)\) is:
- A. 4 V/m
- B. 6 V/m
- C. \(2\sqrt{10}\) V/m
- D. 10 V/m
Solution
The electric field is the negative gradient of potential:
\[
E_x = -\frac{\partial V}{\partial x} = -6x \quad \Rightarrow \quad E_x(1, 2) = -6\,\text{V/m}
\]
\[
E_y = -\frac{\partial V}{\partial y} = -4 \quad \Rightarrow \quad E_y(1, 2) = -4\,\text{V/m}
\]
Magnitude of electric field:
\[
|E| = \sqrt{E_x^2 + E_y^2} = \sqrt{36 + 16} = \sqrt{40} = 2\sqrt{10}\,\text{V/m}
\]
So the correct option is C.