A parallel plate capacitor has circular plates each of radius 5.0 cm. It is being charged so that electric field in the gap between its plates rises steadily at the rate of `2 xx 10^(12) Vm^(-1)s^(-1)`What is the displacement current.

Answer

To solve the problem of finding the displacement current in a parallel plate capacitor with circular plates, we can follow these steps:

### Step 1: Identify the given values
- Radius of the circular plates, \( r = 5.0 \, \text{cm} = 0.05 \, \text{m} \)
- Rate of change of electric field, \( \frac{dE}{dt} = 2 \times 10^{12} \, \text{V/m/s} \)

### Step 2: Calculate the area of the circular plates
The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (0.05)^2 = \pi (0.0025) \approx 7.85 \times 10^{-3} \, \text{m}^2 \]

### Step 3: Use the formula for displacement current
The displacement current \( I_D \) can be calculated using the formula:
\[ I_D = \epsilon_0 \frac{d\Phi}{dt} \]
where \( \Phi \) is the electric flux, and is given by:
\[ \Phi = E \cdot A \]
Thus, the rate of change of electric flux is:
\[ \frac{d\Phi}{dt} = A \frac{dE}{dt} \]
Substituting this into the displacement current formula:
\[ I_D = \epsilon_0 A \frac{dE}{dt} \]

### Step 4: Substitute the known values
The permittivity of free space \( \epsilon_0 \) is approximately:
\[ \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \]
Now substituting the values into the equation:
\[ I_D = (8.85 \times 10^{-12}) \times (7.85 \times 10^{-3}) \times (2 \times 10^{12}) \]

### Step 5: Calculate the displacement current
Calculating the above expression:
\[ I_D = (8.85 \times 10^{-12}) \times (7.85 \times 10^{-3}) \times (2 \times 10^{12}) \]
\[ I_D = (8.85 \times 7.85 \times 2) \times (10^{-12} \times 10^{-3} \times 10^{12}) \]
\[ I_D = (138.61) \times (10^{-3}) \approx 0.13861 \, \text{A} \]

### Step 6: Round to appropriate significant figures
Thus, rounding to two decimal places, we find:
\[ I_D \approx 0.14 \, \text{A} \]

### Final Answer
The displacement current \( I_D \) is approximately \( 0.14 \, \text{A} \).

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