Consider the following. (Let C1 = 20.80 µF and C2 = 14.80 µF.) A rectangular circuit contains a battery and four capacitors. The
bottom side has a 9.00 V battery with the positive terminal on the left. The left and right sides of the circuit each contain a capacitor labeled C1. The top side splits into two parallel horizontal branches, which recombine before reaching the top right corner. There is a 6.00 µF capacitor on the upper branch and a capacitor labeled C2 on the lower branch.
(a) Find the equivalent capacitance of the capacitors in the figure. µF
(b) Find the charge on each capacitor. on the right 20.80 µF capacitor µC on the left 20.80 µF capacitor µC on the 14.80 µF capacitor µC on the 6.00 µF capacitor µC
(c) Find the potential difference across each capacitor. on the right 20.80 µF capacitor V on the left 20.80 µF capacitor V on the 14.80 µF capacitor V on the 6.00 µF capacitor V
(a) Find the equivalent capacitance of the capacitors in the figure. µF
(b) Find the charge on each capacitor. on the right 20.80 µF capacitor µC on the left 20.80 µF capacitor µC on the 14.80 µF capacitor µC on the 6.00 µF capacitor µC
(c) Find the potential difference across each capacitor. on the right 20.80 µF capacitor V on the left 20.80 µF capacitor V on the 14.80 µF capacitor V on the 6.00 µF capacitor V
1 answer:
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Answer:
Explanation:
Given that,
Number of extra electrons, n = 21749
We need to find the net charge on the metal ball. Let Q is the net charge.
We know that the charge on an electron is
To find the net charge if there are n number of extra electrons is :
Q = n × q
So, the net charge on the metal ball is . Hence, this is the required solution.