Electrostatics

JEE Main level, single correct, medium

148 of 311

Question 148 of 311

The electric potential as a function of x,yx, y is given by V=5(x2−y2)V = 5(x^2 - y^2) V. The electric field at a point (2,3)(2, 3) m is __________ V/m.

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Answer: (A) (−20i^+30j^)(-20\hat{i} + 30\hat{j})

The electric field E⃗\vec{E} is given by the negative gradient of the electric potential VV:

E⃗=−(∂V∂xi^+∂V∂yj^)\vec{E} = -\left( \dfrac{\partial V}{\partial x} \hat{i} + \dfrac{\partial V}{\partial y} \hat{j} \right)

Given V=5(x2−y2)V = 5(x^2 - y^2), we find the partial derivatives:

∂V∂x=10x\dfrac{\partial V}{\partial x} = 10x

∂V∂y=−10y\dfrac{\partial V}{\partial y} = -10y

Substituting these into the expression for E⃗\vec{E}:

E⃗=−(10xi^−10yj^)=−10xi^+10yj^\vec{E} = -(10x \hat{i} - 10y \hat{j}) = -10x \hat{i} + 10y \hat{j}

At the point (2,3)(2, 3), substitute x=2x = 2 and y=3y = 3:

E⃗=−10(2)i^+10(3)j^=−20i^+30j^\vec{E} = -10(2) \hat{i} + 10(3) \hat{j} = -20\hat{i} + 30\hat{j} V/m

Answer: (−20i^+30j^)(-20\hat{i} + 30\hat{j})

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