The electric field E is given by the negative gradient of the electric potential V:
E=−(∂x∂Vi^+∂y∂Vj^)
Given V=5(x2−y2), we find the partial derivatives:
∂x∂V=10x
∂y∂V=−10y
Substituting these into the expression for E:
E=−(10xi^−10yj^)=−10xi^+10yj^
At the point (2,3), substitute x=2 and y=3:
E=−10(2)i^+10(3)j^=−20i^+30j^ V/m
Answer: (−20i^+30j^)