# Determining Electric Field on a Charge Plane

Question:
A plane $$\x+2y=5\$$ carries charge $$\\rho_s=6nC/m^2\$$. Determine electric field at $$\(-1,0,1)\$$.
My try:
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.

• 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. Commented Dec 10, 2020 at 9:13
• @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. Commented Dec 10, 2020 at 9:27
• I think your calculation of normal vector is wrong ,it should be -ve because electric field equal to negative of gradient of scalar potential Commented Dec 10, 2020 at 9:50