The electric potential at a point is space is given by the relation V = 23x - 27y + 11y² + 72.What is the magnitude of the force on a 3.4 $\mu$C charge at the point r = (3.0,2.0,1.0)?

Question

Quizplus · Accepted Answer

To find the force on the charge, we need to calculate the electric field at the point r and then use the equation F=qE, where q is the charge. The electric field E is given by the negative gradient of the potential: E = -∇V = (-∂V/∂x, -∂V/∂y, -∂V/∂z). 
Putting the given values, E=(23, -27+11y, 0). At the point r=(3.0, 2.0, 1.0), y=2.0, so E=(23, -5, 0). 
Now, the force F=qE=(3.4e-6)*(23, -5, 0)=(7.82e-5, -1.7e-5, 0). 
So, the magnitude of force |F|=sqrt((7.82e-5)^2 + (-1.7e-5)^2 + 0^2)=9.7e-5 N ≈ 97 μN. 
Hence, the correct answer is choice (C).

The Electric Potential at a Point Is Space Is Given μ\muμ

The Electric Potential at a Point Is Space Is Given $\mu$