a metal ball with a charge of 5.4x10^8 electrons is touched to another metal ball so that all the excess electrons are shared equally. What is the final cahrge on the first ball?
5.4•10^8/2 =2.7•10^8.
q = N•e= 2.7•10^8•1.6•10^-19 =
=4.33•10^-11 C
To find the final charge on the first metal ball, we can use the principle of charge conservation. When two objects with different charges are touched, their charges are shared or redistributed until they reach an equilibrium. In this case, the excess electrons from the first ball are shared equally with the second ball.
Let's first calculate the total charge of the excess electrons on the first metal ball:
Total charge = Number of electrons × Charge of each electron
Given that the number of excess electrons on the first ball is 5.4 × 10^8, we can use the elementary charge (e) as an approximation for the charge of each electron, which is equal to 1.6 × 10^-19 C.
Total charge = 5.4 × 10^8 electrons × (1.6 × 10^-19 C/electron)
= 8.64 × 10^-11 C
Since the excess electrons are shared equally between the two balls, each ball receives half of the total charge. Therefore, the final charge on the first metal ball is half of the calculated total charge.
Final charge on the first ball = 8.64 × 10^-11 C ÷ 2
= 4.32 × 10^-11 C
Hence, the final charge on the first ball after sharing the excess electrons equally with the second ball is 4.32 × 10^-11 C.