The value of the second charge is 1.2 nC.
<h3>
Electric potential</h3>
The work done in moving the charge from infinity to the given position is calculated as follows;
W = Eq₂
E = W/q₂
<h3>Magnitude of second charge</h3>
The magnitude of the second charge is determined by applying Coulomb's law.

Thus, the value of the second charge is 1.2 nC.
Learn more about electric potential here: brainly.com/question/14306881
Answer:
option D
Explanation:
given,
Intensity of sound = 20 dB
distance = 15 m
intensity of sound is increased to = 50 dB
distance between the sound level = ?
Using relation

L₁ = 20 dB L₂ = 50 dB r₁ = 15 m r₂ = ?





r₂ = 0.47 m
r₂ = 47 cm
hence, the correct answer is option D
This is possible due to self-discharge. Very small internal currents inevitably occur in these cells over time and they will eventually exhaust the chemistry.