Answer:
If I double the current in the inductor, the new total energy will become 4E (option f).
Explanation:
The coil or inductor is a passive component made of an insulated wire that stores energy in the form of a magnetic field due to its form of coiled turns of wire, through a phenomenon called self-induction. In other words, inductors store energy in the form of a magnetic field. The energy stored in the space where there is a magnetic field in the inductor is:

where E is Energy [J], L is Inductance [H] and I is Current [A].
If you double the current in the inductor, then the new value of the current is I'= 2*I. So replacing the new total energy is:

Then:

<em><u>If I double the current in the inductor, the new total energy will become 4E (option f).</u></em>
IT IS EASIER TO CLIMB A SLANTED SLOPE
Complete question:
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.
Answer:
The value of its capacitance is 1.027 x 10⁻¹² F
Explanation:
Given;
area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²
separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m
voltage of the battery, V = 18 V
The value of its capacitance is calculated as;

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F
ytyjjryjtjtyjtyjtyyjetyeyjetyyjeytjtyjetyjtjetyjetyeyjetyj
38*10=380 N
To be more exact, 38 should be multiplied by 9.8 instead of 10.