AP Physics question

Test Preparation
1

<p>Here is a problem that seems to show up quite frequently:</p>

<p>Two parallel conducting plates, separated by a distance d, are connected to a battery of emf E. Which of the following is correct if the plate separation is doubled while the battery remains connected?</p>

<p><a href="A">b</a>** The electric charge on the plates is doubled
<a href="B">b</a>** The electric charge on the plates is halved
<a href="C">b</a>** The potential difference between the plates is doubled
<a href="D">b</a>** The potential difference between the plates is halved
<a href="E">b</a>** The capacitance is unchanged</p>

<p>Thanks!</p>

2

<p>Capacitance=(Dielectric constant<em>Vacuum permittivity constant</em>Area)/(Distance)</p>

<p>If distance, if doubled, original capacitance is multiplied by 1/2. So Q/V is halved. Since the battery stays connected, V is the same. Therefore, Q is halved. Answer is B.</p>

3

<p>Correct.</p>

<p>Lets say the battery was disconnected and then the distance between the plates was doubled and then the battery was reconnected. What would the answer be then?</p>

<p>(instead of V being the same, Q would be the same, right?)</p>