The energy and capacity of a charged parallel plate capacitor are E and C respectively. Now a dielective slab of $\epsilon_r=6$is inserted in it then energy and capacity becomes (Assuming charge on plates remains constant)

Question

( A ) $E, 6C$
( B ) $E, C$
( C ) $\frac{E}{6},6C$
( D ) $6E, 6C$