Given a parallel plate capacitor of capacity `C = 80 xx 10^(- 12)` F and is connected to a voltage source. If the current flowing is fount to be I = 0 .15 A, find the potential difference between the plates :

Options

A. `2.41 xx 10^9 V s^(- 1)`
B. `1.87 xx 10^9 v s^(- 1)`
C. `3.2 xx 10^(-4) V s^(-1)`
D. `4.5 xx 10^(-4) V s^(-1)`

Explanation

To find the potential difference between the plates of a parallel plate capacitor given its capacitance and the current flowing through it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between charge, capacitance, and voltage**:
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \cdot V \]
where \( C \) is the capacitance and \( V \) is the potential difference.

2. **Differentiate the equation with respect to time**:
By differentiating both sides of the equation \( Q = C \cdot V \) with respect to time \( t \), we get:
\[ \frac{dQ}{dt} = C \cdot \frac{dV}{dt} \]
Here, \( \frac{dQ}{dt} \) represents the current \( I \).

3. **Substitute the known values**:
We know that:
\[ I = C \cdot \frac{dV}{dt} \]
Rearranging gives us:
\[ \frac{dV}{dt} = \frac{I}{C} \]

4. **Plug in the values**:
Given:
- Capacitance \( C = 80 \times 10^{-12} \) F
- Current \( I = 0.15 \) A

Substituting these values into the equation:
\[ \frac{dV}{dt} = \frac{0.15}{80 \times 10^{-12}} \]

5. **Calculate \( \frac{dV}{dt} \)**:
Performing the division:
\[ \frac{dV}{dt} = \frac{0.15}{80 \times 10^{-12}} = \frac{0.15}{80} \times 10^{12} \]
\[ = 0.001875 \times 10^{12} \]
\[ = 1.875 \times 10^{9} \text{ V/s} \]

6. **Final answer**:
Therefore, the potential difference between the plates is:
\[ \frac{dV}{dt} = 1.87 \times 10^{9} \text{ V/s} \]