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@@ -205,12 +205,12 @@
     \[C_\text{total} = \frac{Q_\text{total}}{V_\text{total}}\]
     Notice that $V_\text{total}$ is the sum of the voltages across $C_1$ and $C_2$. We can get each of these voltages using the definition of capacitance.
     \[V_\text{total} = V_1 + V_2 = \frac{Q_1}{C_1} + \frac{Q_2}{C_2}\]
-    The key observation now is that because the right plate of $C_1$ is connected to the left plate of $C_2$, the charge stored on both plates must be equal. \footnotemark Therefore, we have $Q_1 = Q_2$. Let us call this charge stored $Q$ (i.e. $Q = Q_1 = Q_2$). Now, we know that the total charge stored is also $Q$. \footnotemark Therefore, we know that $Q_\text{total} = Q$. Now, we have 
+    The key observation now is that because the right plate of $C_1$ is connected to the left plate of $C_2$, the charge stored on both plates must be of equal magnitude.\footnotemark Therefore, we have $Q_1 = Q_2$. Let us call this charge stored $Q$ (i.e. $Q = Q_1 = Q_2$). Now, we know that the total charge stored is also $Q$.\footnotemark Therefore, we know that $Q_\text{total} = Q$. Now, we have 
     \[C_\text{total} = \frac{Q_\text{total}}{V_\text{total}} = \frac{Q}{Q_1/C_1 + Q_2/C_2} = \frac{Q}{Q/C_1 + Q/C_2} = \frac{1}{1/C_1 + 1/C_2}\]
 
-    \footnotetext{If this were not true, then there would a difference of charge between these two plates which means there would be a voltage across these two connected plates. This violates the idea that two points that are electrically connected must have the same voltage.}
+    \footnotetext{If this were not true, then there would be a net charge on these two plates and the wire between them. Because we assume that the capacitors started out with no net charge and there is no way for charge to leave the middle wire or the two plates it connects, this is impossible. }
 
-    \footnotetext{If you are having trouble seeing this, suppose we apply a positive voltage to the left plate of $C_1$ relative to the right plate of $C_2$. Suppose this causes the left plate of $C_1$ to charge to some charge $q$. Thus we must have the right plate of $C_1$ has a charge of $-q$ by pushing $q$ units of charge onto the left plate of $C_2$. Now the left of $C_2$ has $q$ units of charge which causes a corresponding $-q$ charge on the right side of $C_2$. Thus the overall total charge separated across these two capacitors is $q$.
+    \footnotetext{If you are having trouble seeing this, suppose we apply a positive voltage to the left plate of $C_1$ relative to the right plate of $C_2$. Suppose this causes the left plate of $C_1$ to charge to some charge $q$. We now must have a charge of $-q$ on the right plate of $C_1$ because $q$ units of charge are now pushed onto the left plate of $C_2$. Now the left of $C_2$ has $q$ units of charge which causes a corresponding $-q$ charge on the right side of $C_2$. Thus the overall total charge separated across these two capacitors is $q$. 
     }
 
     \ex{1.16}