A plane \$x+2y=5\$ carries charge \$\rho_s=6nC/m^2\$. Determine electric field at \$(-1,0,1)\$.
My try:
At first, I calculated gradient
Let, \$\phi=x+2y-5\$
\$\begin{align}\\ \vec\nabla\phi&=\frac{\partial\phi}{\partial x}\hat a_x+\frac{\partial\phi}{\partial y}\hat a_y+\frac{\partial\phi}{\partial z}\hat a_z\\ &=\hat a_x+2\hat a_y \end{align}\\\$
We know, \$\vec E=\frac{\rho_s}{2\epsilon_0}\hat a_n\$, where \$\hat a_n\$ is the normal vector to the surface
Here, \$\hat a_n=\frac{\hat a_x+2\hat a_y}{\sqrt{1^2+2^2}}=\frac{1}{\sqrt{5}}(\hat a_x+2\hat a_y)\$
\$\begin{align}\\ \therefore\vec E&=\frac{6\times 10^{-9}}{2\times\epsilon_0}.\frac{1}{\sqrt{5}}(\hat a_x+2\hat a_y)\\ &=151.53\hat a_x+303.1\hat a_y\\ \end{align}\\\$
But, in my book answer is:\$=-151.53\hat a_x-303.1\hat a_y\$. I can't understand why there is minus sign. Please anyone check this.

  • \$\begingroup\$ Well done, all the digits are correct. The sign is wrong, that's only a sign convention. Sign inversion can hide anywhere, especially when computing a plane normal. For anybody to help you, you'll have to spell out the conventions you've used at every step. I've a feeling that by the time you've done that. you'll have spotted your difference to the book. \$\endgroup\$
    – Neil_UK
    Commented Dec 10, 2020 at 9:13
  • \$\begingroup\$ @Neil_UK my problem is I can't draw the diagram. So, I am unsure about my answer. I guess \$x+2y=5\$ is at z=0, so I assumed (-1,0,1) to be above the plane. That's why I took the positive sign. \$\endgroup\$
    – Ankita Pal
    Commented Dec 10, 2020 at 9:27
  • 1
    \$\begingroup\$ I think your calculation of normal vector is wrong ,it should be -ve because electric field equal to negative of gradient of scalar potential \$\endgroup\$
    – user215805
    Commented Dec 10, 2020 at 9:50

1 Answer 1


because normal vector towards negative from line to <-1,0,1> this point is located inwards relative to the line


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