# Inserting Dielectric into a capacitor in parallel

1. Feb 14, 2015

### Rabbittt

1. The problem statement, all variables and given/known data
Two identical capacitors Ca and Cb each of capacitance 6.60 http://lon-capa.mines.edu/adm/jsMath/fonts/cmmi10/alpha/100/char16.png F are connected in parallel across a CONSTANT total electric potential difference of 480 V. A dielectric slab of dielectric constant 4.65 can fill Ca and is slowly inserted into that capacitor.

What is the change in charge, [PLAIN]http://lon-capa.mines.edu/adm/jsMath/fonts/cmr10/alpha/100/char01.pngQa, [Broken] on Ca when the dielectric is added to Ca?

2. Relevant equations
C=Q/V

3. The attempt at a solution
Ca+Cb= Qtotal/480
Qtotal/2=Qa initial
because they have the same capacitance
Ca=kQa final/480

Qa final- Qa initial = Change in Qa

Any help would be greatly appreciated.

Last edited by a moderator: May 7, 2017
2. Feb 14, 2015

### Staff: Mentor

Hi Rabbitt, Welcome to Physics Forums.

Since the capacitors are in parallel across the voltage source you can ignore Cb and just deal with Ca alone. This is true because components in parallel have the same potential difference; it's as though they each have their own separate voltage source and operate independently.

A dielectric affects the capacitance of a capacitor (in what way?), so why not start by determining the new capacitor value for Ca?