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Question: The electric potential at point (1, 3) of an electrostatic field which changes with co-ordinates as ...

The electric potential at point (1, 3) of an electrostatic field which changes with co-ordinates as E⃗=6xyi^+3(x2−y2)j^\vec{E}=6xy\hat{i}+3(x^2-y^2)\hat{j} taking potential at origin is 0 V.

Answer

18 V

Explanation

Solution

The electric potential VV is related to the electric field E⃗\vec{E} by the equation E⃗=−∇V\vec{E} = -\nabla V.
Given the electric field E⃗=6xyi^+3(x2−y2)j^\vec{E}=6xy\hat{i}+3(x^2-y^2)\hat{j}, we can write its components as:
Ex=6xyE_x = 6xy
Ey=3(x2−y2)E_y = 3(x^2-y^2)

From the relationship E⃗=−∇V\vec{E} = -\nabla V, we have:
Ex=−∂V∂x  ⟹  ∂V∂x=−6xy(1)E_x = -\frac{\partial V}{\partial x} \implies \frac{\partial V}{\partial x} = -6xy \quad (1)
Ey=−∂V∂y  ⟹  ∂V∂y=−3(x2−y2)=−3x2+3y2(2)E_y = -\frac{\partial V}{\partial y} \implies \frac{\partial V}{\partial y} = -3(x^2-y^2) = -3x^2 + 3y^2 \quad (2)

First, let's check if the field is conservative by verifying if ∇×E⃗=0\nabla \times \vec{E} = 0. For a 2D field, this means ∂Ey∂x−∂Ex∂y=0\frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} = 0.
∂Ey∂x=∂∂x(3x2−3y2)=6x\frac{\partial E_y}{\partial x} = \frac{\partial}{\partial x} (3x^2 - 3y^2) = 6x
∂Ex∂y=∂∂y(6xy)=6x\frac{\partial E_x}{\partial y} = \frac{\partial}{\partial y} (6xy) = 6x
Since ∂Ey∂x=∂Ex∂y\frac{\partial E_y}{\partial x} = \frac{\partial E_x}{\partial y}, the field is conservative, and a potential function V(x,y)V(x,y) exists.

Now, integrate equation (1) with respect to xx, treating yy as a constant:
V(x,y)=∫(−6xy)dx=−6y∫xdx=−6y(x22)+f(y)V(x, y) = \int (-6xy) dx = -6y \int x dx = -6y \left(\frac{x^2}{2}\right) + f(y)
V(x,y)=−3x2y+f(y)(3)V(x, y) = -3x^2y + f(y) \quad (3)
where f(y)f(y) is an arbitrary function of yy.

Next, differentiate equation (3) with respect to yy and compare it with equation (2):
∂V∂y=∂∂y(−3x2y+f(y))=−3x2+dfdy\frac{\partial V}{\partial y} = \frac{\partial}{\partial y} (-3x^2y + f(y)) = -3x^2 + \frac{df}{dy}
Comparing this with equation (2), which states ∂V∂y=−3x2+3y2\frac{\partial V}{\partial y} = -3x^2 + 3y^2:
−3x2+dfdy=−3x2+3y2-3x^2 + \frac{df}{dy} = -3x^2 + 3y^2
dfdy=3y2\frac{df}{dy} = 3y^2

Now, integrate this expression with respect to yy to find f(y)f(y):
f(y)=∫3y2dy=3(y33)+C=y3+Cf(y) = \int 3y^2 dy = 3\left(\frac{y^3}{3}\right) + C = y^3 + C
where CC is the integration constant.

Substitute f(y)f(y) back into equation (3) to get the complete potential function:
V(x,y)=−3x2y+y3+CV(x, y) = -3x^2y + y^3 + C

We are given that the potential at the origin (0, 0) is 0 V. Use this condition to find CC:
V(0,0)=−3(0)2(0)+(0)3+C=0+0+C=CV(0, 0) = -3(0)^2(0) + (0)^3 + C = 0 + 0 + C = C
Since V(0,0)=0V(0, 0) = 0, we have C=0C = 0.

So, the electric potential function is:
V(x,y)=−3x2y+y3V(x, y) = -3x^2y + y^3

Finally, we need to find the electric potential at point (1, 3). Substitute x=1x=1 and y=3y=3 into the potential function:
V(1,3)=−3(1)2(3)+(3)3V(1, 3) = -3(1)^2(3) + (3)^3
V(1,3)=−3(1)(3)+27V(1, 3) = -3(1)(3) + 27
V(1,3)=−9+27V(1, 3) = -9 + 27
V(1,3)=18V(1, 3) = 18 V