Electrostatics

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Question 157 of 311

Electric field in a region is given by E⃗=Axi^+Byj^\vec{E}=A x \hat{i}+B y \hat{j}, where A=10 V/m2A=10 \mathrm{~V} / \mathrm{m}^{2} and B=5 V/m2B=5 \mathrm{~V} / \mathrm{m}^{2}. If the electric potential at a point (10,20)(10,20) is 500 V, then the electric potential at origin is ____\_\_\_\_ V.

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Answer: (B) 2000

The electric field is related to potential by E⃗=−∇V\vec{E} = -\nabla V.
Given: E⃗=Axi^+Byj^\vec{E} = Ax\hat{i} + By\hat{j} with A=10A = 10 V/m² and B=5B = 5 V/m²
This means:
∂V∂x=−Ax\frac{\partial V}{\partial x} = -Ax and ∂V∂y=−By\frac{\partial V}{\partial y} = -By
Integrating:
V(x,y)=−Ax22−By22+C=−5x2−5y22+CV(x,y) = -\frac{Ax^2}{2} - \frac{By^2}{2} + C = -5x^2 - \frac{5y^2}{2} + C
Using the boundary condition at (10,20)(10, 20) where V=500V = 500 V:
500=−5(100)−5(400)2+C500 = -5(100) - \frac{5(400)}{2} + C
500=−500−1000+C500 = -500 - 1000 + C
C=2000C = 2000 V
At the origin (0,0)(0, 0):
V(0,0)=−5(0)−5(0)2+2000=2000V(0,0) = -5(0) - \frac{5(0)}{2} + 2000 = 2000 V

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