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JEE MAIN_2026
paper 1
Question
The dimensional formula of $\frac{1}{2} \epsilon_0 \mathrm{E}^2$ ( $\epsilon_0=$ permittivity of vacuum and $\mathrm{E}=$ electric field) is $\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}$. The value of $2 a-b+c=$ $\_\_\_\_$ .
Select the correct option:
A
0.0
B
1
C
-1
D
2
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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