Electric displacement is given by `D=epsilonE,` here, `epsilon=` electric permittivity, E = electric field strength
The dimensions of electric displacement are

Options

A. `[ML^(-2)TA]`
B. `[L^(-2)T^(-1)A]`
C. `[L^(-2)TA]`
D. None of these

Answer

To find the dimensions of electric displacement \( D \), given by the equation \( D = \epsilon E \), we need to determine the dimensions of both \( \epsilon \) (electric permittivity) and \( E \) (electric field strength).

### Step-by-Step Solution:

1. **Understanding Electric Field Strength \( E \)**:
- The electric field strength \( E \) is defined as the force \( F \) per unit charge \( Q \):
\[ E = \frac{F}{Q} \]
- The dimensions of force \( F \) are given by \( [F] = [M][L][T^{-2}] \) (mass × acceleration).
- The dimensions of charge \( Q \) are represented as \( [Q] = [I][T] \) (current × time).
- Therefore, the dimensions of electric field \( E \) can be expressed as:
\[ [E] = \frac{[F]}{[Q]} = \frac{[M][L][T^{-2}]}{[I][T]} = [M][L][T^{-3}][I^{-1}] \]

2. **Understanding Electric Permittivity \( \epsilon \)**:
- The relationship involving electric permittivity can be derived from Coulomb's law:
\[ F = \frac{1}{4 \pi \epsilon} \frac{Q_1 Q_2}{r^2} \]
- Rearranging gives:
\[ \epsilon = \frac{Q_1 Q_2}{4 \pi F r^2} \]
- The dimensions of \( \epsilon \) can be derived as follows:
- The dimensions of \( Q^2 \) are \( [Q^2] = [I^2][T^2] \).
- The dimensions of \( r^2 \) are \( [L^2] \).
- Thus, the dimensions of \( \epsilon \) are:
\[ [\epsilon] = \frac{[Q^2]}{[F][L^2]} = \frac{[I^2][T^2]}{[M][L][T^{-2}][L^2]} = \frac{[I^2][T^2]}{[M][L^3][T^{-2}]} = \frac{[I^2][T^4]}{[M][L^3]} \]

3. **Finding Dimensions of Electric Displacement \( D \)**:
- Now, substituting the dimensions of \( \epsilon \) and \( E \) into the equation \( D = \epsilon E \):
\[ [D] = [\epsilon][E] = \left(\frac{[I^2][T^4]}{[M][L^3]}\right) \left([M][L][T^{-3}][I^{-1}]\right) \]
- Simplifying this gives:
\[ [D] = \frac{[I^2][T^4]}{[M][L^3]} \cdot [M][L][T^{-3}][I^{-1}] = \frac{[I^2][T^4][M][L]}{[M][L^3][I]} = \frac{[I][T^4]}{[L^2]} \]
- Therefore, the final dimensions of electric displacement \( D \) are:
\[ [D] = [I][L^{-2}][T] \]

### Final Answer:
The dimensions of electric displacement \( D \) are given by:
\[ [D] = [I][L^{-2}][T] \]