Answer: A) 2 B) 4 C) 1
Explanation:
The Electric field from a parallel-plate capacitor is given by:
A) E=Q/(L^2 * ε0) so if we put a charge double the final electric field is double that the original.
B) from the above expression for the electric field, If the size of the plate is double, then the E final is four times weaker that the original.
C) If the distante between plates is doubled the final electric field is the same that initial.
