Q83P An isolated large plate capacito... [FREE SOLUTION]

Chapter 16: Q83P (page 671)

An isolated large plate capacitor (not connected to anything) originally has a potential difference of 1000 V with an air gap of 2 mm. Then a plastic slab 1 mm thick, with dielectric constant 5, is inserted in the middle of the air gap as shown in figure 16.96. Calculate the following potential differences and explain your work.
$\begin{array}{l} V_{1} - V_{2} = ? V_{2} - V_{3} = ? \\ V_{3} - V_{4} = ? V_{1} - V_{4} = ? \end{array}$

Short Answer

Expert verified

The potential difference between point 1 and 2 is 250 V, between point 2 and 3 is 100V , between point 3 and 4 is 250V and between point 1 and 4 is 600V .

Step by step solution

Identification of given data

The distance between plates of capacitor is l = 2 mm .

The potential difference across plates of capacitor is V = 1000 V .

The distance between point 1 and 2, point 3 and 4 is d = 0.5 mm

The thickness of plastic slab is t = 1 mm

The dielectric constant of plastic slab is k = 5

Conceptual Explanation

The potential difference for the air gap is calculated by the electric field multiplied by air gap. The potential difference for plastic slab is found by assuming slab dielectric material.

Determination of potential differences between given points

The electric field for air gap between plates is given as:

$E = \frac{V}{I} E = \frac{1000 V}{(2 m m) (\frac{1 m}{1000 m m})} E = 5 脳 10^{5} V / m$

The electric field for plastic slab between plates is given as:

$E_{s} = \frac{E}{K} E_{s} = \frac{5 脳 10^{5} V / m}{5} E_{s} = 1 脳 10^{5} V / m$

The electric potential between point 1 and point 2 is given as:

$V_{2} - V_{1} = E . d V_{2} - V_{1} = (5 脳 10^{5} V / m) (0.5 m m) (\frac{1 m}{1000 m m}) V_{2} - V_{1} = 250 V$

The electric potential between point 2 and point 3 is given as:

$V_{2} - V_{3} = E_{s} . t V_{2} - V_{3} = (1 脳 10^{5} V / m) (1 m m) (\frac{1 m}{1000 m}) V_{2} - V_{3} = 100 V$

The electric potential between point 3 and point 4 is given as:

$V_{3} - V_{4} = E_{s} . d V_{3} - V_{4} = (1 脳 10^{5} V / m) (0.5 m m) (\frac{1 m}{1000 m}) V_{3} - V_{4} = 250 V$

The electric potential between point 1 and point 4 is given as:

$\begin{array}{l} V_{1} - V_{4} = (V_{1} - V_{2}) + (V_{2} - V_{3}) + (V_{3} - V_{4}) \\ V_{1} - V_{4} = 250 V + 100 V + 250 V \\ V_{1} - V_{4} = 600 V \end{array}$

Therefore, the potential difference between point 1 and 2 is 250V , between point 2 and 3 is 100V , between point 3 and 4 is 250V and between point 1 and 4 is 600V.