Given that `C` denotes capacitance of a capacitor and `V` is the potential difference across its plates. Then the dimensions of `CV^(2)` are same as that of

Options

A. momentum
B. power
C. forces
D. torque

Answer

To solve the problem, we need to determine the dimensions of the expression \( CV^2 \) where \( C \) is the capacitance and \( V \) is the potential difference across the plates of a capacitor.

### Step-by-step Solution:

1. **Identify the dimensions of capacitance \( C \)**:
The capacitance \( C \) is defined as the charge \( Q \) per unit potential difference \( V \):
\[ C = \frac{Q}{V} \]
The dimension of charge \( Q \) is \( [I][T] \) (where \( I \) is current and \( T \) is time). The potential difference \( V \) has the dimension of energy per unit charge, which can be expressed as:
\[ V = \frac{E}{Q} \]
The dimension of energy \( E \) is \( [M][L^2][T^{-2}] \). Therefore, the dimension of \( V \) is:
\[ [V] = \frac{[M][L^2][T^{-2}]}{[I][T]} = [M][L^2][T^{-3}][I^{-1}] \]

2. **Substituting the dimensions into the capacitance equation**:
Now substituting the dimensions of \( Q \) and \( V \) into the capacitance equation:
\[ [C] = \frac{[I][T]}{[M][L^2][T^{-3}][I^{-1}]} = \frac{[I^2][T^4]}{[M][L^2]} \]

3. **Calculate the dimensions of \( CV^2 \)**:
Now we need to calculate the dimensions of \( CV^2 \):
\[ CV^2 = C \cdot V \cdot V \]
The dimension of \( V^2 \) is:
\[ [V^2] = \left([M][L^2][T^{-3}][I^{-1}]\right)^2 = [M^2][L^4][T^{-6}][I^{-2}] \]
Therefore, the dimensions of \( CV^2 \) are:
\[ [CV^2] = [C] \cdot [V^2] = \left(\frac{[I^2][T^4]}{[M][L^2]}\right) \cdot \left([M^2][L^4][T^{-6}][I^{-2}]\right) \]
Simplifying this:
\[ [CV^2] = \frac{[I^2][T^4][M^2][L^4][T^{-6}][I^{-2}]}{[M][L^2]} = [M][L^2][T^{-2}] \]

4. **Conclusion**:
The dimensions of \( CV^2 \) are the same as the dimensions of energy, which is \( [M][L^2][T^{-2}] \).

### Final Answer:
The dimensions of \( CV^2 \) are the same as that of energy.

---