Question

an uncharged parallel plate capacitor having dielectric constant k is connected to a similar air filled capacitor charge to a potential V .the two capacitor share the charge and the common potential is V1.the dielectric constant k is...

Answer 1

Answer on Question 51080, Physics, Mechanics | Kinematics | Dynamics

Question:

An uncharged parallel-plate capacitor having a dielectric of constant $K$ is connected to a similar air filled capacitor charged to a potential $V$. The two capacitor share the charge and the common potential is $V'$. The dielectric constant is:

1) $\frac{V' - V}{V' + V}$

2) $\frac{V' - V}{V'}$

3) $\frac{V' - V}{V}$

4) $\frac{V - V'}{V'}$

Solution:

By the definition of the capacitance of a parallel-plate capacitor we have:

$$C = \frac{q}{V},$$

where $C$ is the capacitance of a parallel-plate capacitor, $q$ is the charge and $V$ is the voltage between the plates.

Let us write the capacitance and initial charge for both capacitors. The first one is uncharged parallel-plate capacitor having a dielectric of constant $K$, and because of this capacitor is uncharged we can write:

$$C_1 = K C, \quad q_1 = 0.$$

The second one is a similar air filled capacitor charged to a potential $V$, so we can write:

$$C_2 = C, \quad q_2 = C V.$$

Then, from the condition of the question, we know that the two capacitor share the charge and the common potential is $V'$. Thus, we can write the final charges:

$$q_1' = C_1 V = K C V',$$ $$q _ {2} ^ {\prime} = C _ {2} V = C V ^ {\prime}.$$

Therefore, we get:

$$q _ {1} + q _ {2} = q _ {1} ^ {\prime} + q _ {2} ^ {\prime},$$ $$0 + C V = K C V ^ {\prime} + C V ^ {\prime},$$ $$C V = C V ^ {\prime} (K + 1),$$ $$\frac {V}{V ^ {\prime}} = K + 1,$$ $$K = \frac {V}{V ^ {\prime}} - 1 = \frac {V - V ^ {\prime}}{V ^ {\prime}}.$$

Answer:

4) $\frac{V - V^{\prime}}{V^{\prime}}$

http://www.AssignmentExpert.com/