Question

The electric potential at a region in terms of the coordinates \((x,y,z)\) in metre is

\[ V(x,y,z)=\left(10-3x^2-y^2-1.5z^2\right)\times10^6\ \text{V}. \]
The magnitude of the electric force acting on a charge of \(2\,\mu\text{C}\) placed at the point \((1,1,3)\,\text{cm}\) is

• A. \(0.33\,\text{N}\)
• B. \(0.11\,\text{N}\)
• C. \(0.44\,\text{N}\)
• D. \(0.22\,\text{N}\)

Hint:

Electric field from potential: \[ \boxed{ \vec E=-\nabla V } \] and electric force is \[ \boxed{ \vec F=q\vec E. } \]

Answer 1

The Correct Option is D

Step 1: Find the electric field. The electric field is \[ \vec E=-\nabla V. \] Hence, \[ E_x=6x\times10^6, \] \[ E_y=2y\times10^6, \] \[ E_z=3z\times10^6. \] At \[ (x,y,z)=(0.01,0.01,0.03)\ \text{m}, \] \[ E_x=6\times10^4\ \text{N C}^{-1}, \] \[ E_y=2\times10^4\ \text{N C}^{-1}, \] \[ E_z=9\times10^4\ \text{N C}^{-1}. \]

Step 2: Calculate the magnitude of the electric field. \[ E = \sqrt{(6\times10^4)^2+(2\times10^4)^2+(9\times10^4)^2} = 11\times10^4 = 1.1\times10^5\ \text{N C}^{-1}. \]

Step 3: Find the electric force. Given, \[ q=2\times10^{-6}\ \text{C}. \] Therefore, \[ F=qE = 2\times10^{-6}\times1.1\times10^5 = 0.22\ \text{N}. \] Hence, \[ \boxed{0.22\ \text{N}} \] Therefore, \[ \boxed{(D)} \] is the correct answer.