In this answer I showed how to construct an independent family $\mathscr{I}$ of subsets of $\Bbb N$ of cardinality $2^\omega=\mathfrak{c}$. That is, $\mathscr{I}$ has the property that
$$\bigcap_{A\in\mathscr{J}}A\cap\bigcap_{A\in\mathscr{K}}(X\setminus A)\ne\varnothing$$
whenever $\mathscr{J}$ and $\mathscr{K}$ are disjoint finite subsets of $\mathscr{I}$. Let
$$\mathscr{B}=\mathscr{I}\cup\left\{\Bbb N\setminus\bigcap\mathscr{A}:\mathscr{A}\subseteq\mathscr{I}\text{ and }\mathscr{A}\text{ is infinite}\right\}\;;$$
$\mathscr{B}$ has the strong finite intersection property, so it can be extended to an ultrafilter $\mathscr{U}$ on $\Bbb N$. Suppose that $\mathscr{C}$ is a base for $\mathscr{U}$ of cardinality less than $2^\omega$. $\mathscr{I}\subseteq\mathscr{U}$, so for each $I\in\mathscr{I}$ there is some $C_I\in\mathscr{C}$ such that $C_I\subseteq I$, and $|\mathscr{C}|<|\mathscr{I}|$, so there is a $C\in\mathscr{C}$ such that $\{I\in\mathscr{I}:C_I=C\}$ is infinite. Let $A=\bigcap\{I\in\mathscr{I}:C_I=C\}$; then $C\subseteq A$, so $A\in\mathscr{U}$. But $\Bbb N\setminus A\in\mathscr{B}\subseteq\mathscr{U}$ as well, which is impossible. Thus, $\mathscr{U}$ has no base of cardinality less than $2^\omega$. (The existence of such an ultrafilter is due to Pospíšil.)
Let $\mathscr{U}$ be an ultrafilter on $\Bbb N$ having no base of cardinality less than $2^\omega$, and define
$$c:\wp(\Bbb N)\to\{0,1\}:A\mapsto\begin{cases}
1,&\text{if }A\in\mathscr{U}\\
0,&\text{if }A\notin\mathscr{U}\;;
\end{cases}$$
evidently $c$ is a charge. Assume that $\mathscr{F}\subseteq\wp(\Bbb N)$, and $|\mathscr{F}|\lt 2^\omega$, and let
$$\mathscr{B}_0=(\mathscr{F}\cap\mathscr{U})\cup\{\Bbb N\setminus F:F\in\mathscr{F}\setminus\mathscr{U}\}\subseteq\mathscr{U}\;.$$
Let $\mathscr{B}$ be the closure of $\mathscr{B}_0$ under finite intersections; $|\mathscr{B}_0|<2^\omega$, so $|\mathscr{B}|<2^\omega$, and $\mathscr{B}$ is therefore not a base for $\mathscr{U}$. Thus, there is a $U\in\mathscr{U}$ that does not contain any element of $\mathscr{B}$. It follows that $B\cap(\Bbb N\setminus U)\ne\varnothing$ for each $B\in\mathscr{B}$ and hence that $\{\Bbb N\setminus U\}\cup\mathscr{B}$ can be extended to an ultrafilter $\mathscr{V}$ on $\Bbb N$.
Define the charge
$$d:\wp(\Bbb N)\to\{0,1\}:A\mapsto\begin{cases}
1,&\text{if }A\in\mathscr{V}\\
0,&\text{if }A\notin\mathscr{V}\;.
\end{cases}$$
The construction ensures that for each $F\in\mathscr{F}$ we have $F\in\mathscr{V}$ iff $F\in\mathscr{U}$, so $d\upharpoonright\mathscr{F}=c\upharpoonright\mathscr{F}$. However, $c(U)=1\ne 0=d(U)$, so $c\ne d$.
I’ve never gone through the details, but according to Exercise A$10$ in Chapter VIII of Ken Kunen’s Set Theory: An Introduction to Independence Proofs (North Holland, $1980$), it’s consistent that there there be an ultrafilter on $\Bbb N$ with a base of cardinality $\omega_1<2^\omega$. If we take $\mathscr{U}$ to be such an ultrafilter and $\mathscr{F}$ a base for $\mathscr{U}$ of cardinality $\omega_1$, the charge $c$ defined as above from $\mathscr{U}$ cannot be extended to any other charge agreeing with $c$ on $\mathscr{F}$.
