Question

problem 17.49 an uncharged capacitor is connected to a 25 - v battery until it is fully charged, after which it is disconnected from the battery. a slab of paraffin (k = 2.2) is then inserted between the plates. part a what will now be the voltage between the plates? express your answer using two significant figures and include the appropriate units.

Explanation:

Step1: Recall capacitor - voltage formula with dielectric

The charge $Q$ on the capacitor remains constant after disconnection from the battery. The capacitance of a capacitor with a dielectric is $C = KC_0$, where $K$ is the dielectric constant, $C_0$ is the initial capacitance. Also, $Q = C_0V_0=CV$, where $V_0$ is the initial voltage and $V$ is the final voltage.

Step2: Solve for the final voltage

We can rewrite the equation $C_0V_0 = CV$ as $V=\frac{C_0}{C}V_0$. Since $C = KC_0$, then $V=\frac{V_0}{K}$. Given $V_0 = 25\ V$ and $K = 2.2$.
\[V=\frac{25}{2.2}\ V\]
\[V\approx11\ V\]

Answer:

$11\ V$