diff --git a/docs/1b/ece106.md b/docs/1b/ece106.md
index 35ee429..f908718 100644
--- a/docs/1b/ece106.md
+++ b/docs/1b/ece106.md
@@ -317,13 +317,13 @@ At a point $P$, the electrostatic potential $V_p$ or voltage is the work done pe
 
 $$
 V_p=\lim_{q\to 0^+}\frac{W_i}{q} \\
-W_I=\int^p_\infty\vec F_I\bullet \vec{dl}
+W_I=\int^p_\infty\vec F_I\bullet \vec{dl}=\Delta U=QV_p
 $$
 
 Because the desired force acts opposite to the force from the electric field, as long as $\vec E$ is known at each point:
 
 $$
-V_p=-\int^p_\infty\vec E\bullet\vec{dl}
+V_p=-\int^p_\infty\vec E\bullet\vec{dl} \\
 V_p=-\int^p_\infty E\ dr
 $$
 
@@ -337,7 +337,7 @@ Therefore, the potential due to a point charge is equal to:
 
 $$V_p=-\int^p_\infty\frac{kQ}{r^2}dr=\frac{kQ}{r}$$
 
-**Positive** charges naturally move to **lower** potentials ($V$ decreases) while negative charges do the opposite.
+**Positive** charges naturally move to **lower** potentials ($V$ decreases) while negative charges do the opposite. Potential energy always decreases.
 
 In order to calculate the voltage for charge distributions:
 
@@ -348,3 +348,11 @@ $$V_p=-\int^p_\infty\vec E\bullet\vec{dl}$$
 - If the charge is asymmetric:
 
 $$V_p=\int_\text{charge dist}\frac{kdQ}{r}$$
+
+The electric field always points in the direction of **lower** potential, and is equal to the **negative gradient** of potential.
+
+$$\vec E=-\nabla V$$
+
+If $\vec E$ is constant:
+
+$$\vec E=\frac{Q_{enc\ net}}{\epsilon_0\oint dS}$$