Answer:
option A
Explanation:
given,
area of the plate = A
distance = d
[tex]U = \dfrac{1}{2}CV^2[/tex]
where
C is the capacitance
V is the potential difference
The capacitance of the parallel plate capacitor, at the beginning, is given by
[tex]C = \dfrac{\epsilon_0 A}{d}[/tex]
where ε₀ is the permittivity of free space,
[tex]U_0 = \dfrac{1}{2}\dfrac{\epsilon_0 A}{d}V^2[/tex]
now, the distance is doubled
d' = 2 d
while the potential difference is kept constant. Therefore, we can calculate the new potential energy:
[tex]U = \dfrac{1}{2}\dfrac{\epsilon_0 A}{d'}V^2[/tex]
[tex]U = \dfrac{1}{2}\dfrac{\epsilon_0 A}{2d}V^2[/tex]
[tex]U = \dfrac{U_0}{2}[/tex]
Hence, the correct answer is option A
