Electric field in a region is given by vec{E} = Axhat{i} + Byhat{j}, where A = 10 V/m^2 and B = 5 V/m^2. If the electric potential at a point (10, 20) is 500 V, then the electric potential at origin is ____

Electric field Ax i + By j potential calculation at origin

Q. Electric field in a region is given by

$\vec{E} = Ax\hat{i} + By\hat{j}$, where $A = 10\ \text{V m}^{-2}$ and $B = 5\ \text{V m}^{-2}$.

If the electric potential at a point $(10, 20)$ is $500\ \text{V}$, then the electric potential at origin is _____ V.

A. 1000

B. 0

C. 2000

D. 500

Correct Answer: 2000

Explanation

The relation between electric field and electric potential is given by:

$$ \vec{E} = - \nabla V $$

This gives the component-wise relations:

$$ E_x = -\frac{\partial V}{\partial x}, \quad E_y = -\frac{\partial V}{\partial y} $$

Given electric field:

$$ E_x = Ax = 10x, \quad E_y = By = 5y $$

Now integrate $E_x$ with respect to $x$:

$$ -\frac{\partial V}{\partial x} = 10x $$

$$ \frac{\partial V}{\partial x} = -10x $$

$$ V = -5x^2 + f(y) $$

Now use $E_y$ to find $f(y)$:

$$ -\frac{\partial V}{\partial y} = 5y $$

$$ \frac{\partial V}{\partial y} = -5y $$

Differentiate $V = -5x^2 + f(y)$ with respect to $y$:

$$ \frac{d f(y)}{dy} = -5y $$

$$ f(y) = -\frac{5}{2}y^2 + C $$

Hence, electric potential function is:

$$ V(x,y) = -5x^2 - \frac{5}{2}y^2 + C $$

Now apply the given condition $V(10,20) = 500$:

$$ 500 = -5(10)^2 - \frac{5}{2}(20)^2 + C $$

$$ 500 = -500 - 1000 + C $$

$$ C = 2000 $$

At origin $(0,0)$:

$$ V(0,0) = C = 2000\ \text{V} $$

Therefore, the electric potential at origin is 2000 V.

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