OnlineCBT
Physics · Electric Charges and Fields
हिन्दीIn an isolated electrostatics setup, two initially neutral insulating bodies \( X \) and \( Y \) are rubbed together, resulting in the transfer of \( 3.125 \times 10^{13} \) electrons from \( X \) to \( Y \). Simultaneously, inside the same isolated system, a neutral particle decays into a proton, an electron, and a neutrino. The net final charge on body \( Y \) and the total net charge of the entire isolated system are, respectively:
Correct answer · Option 1
\( -5\,\mu\text{C} \) and \( 0\,\mu\text{C} \)
Explanation
Body Y gains electrons: \( q_Y = -N e = -(3.125 \times 10^{13}) \times (1.6 \times 10^{-19}\text{ C}) = -5\,\mu\text{C} \). Particle decay produces equal positive (+e) and negative (-e) charges, yielding zero net charge. By charge conservation, the net charge of the isolated system remains \( 0\,\mu\text{C} \).
