Physics, 19.02.2020 06:01 Alysssssssssssa

Capacitors C1=10μF and C2=20μF are each charged to 14 V , then disconnected from the battery without changing the charge on the capacitor plates. The two capacitors are then connected in parallel, with the positive plate of C1connected to the negative plate of C2 and vice versa.

(a) Afterward, what is the charge on each capacitor?
(b) What is the potential difference across each capacitor?

Answers: 2

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Answer from: jes69

a) $Q_1 = 0, Q_2 = 140 \mu C$

b) V_1=0, V_2 = 7V

Explanation:

The charge on each capacitor is given by

Q=CV

where

C is the capacitance

V is the voltage

For the 1st capacitor, we have

$Q_1 = C_1 V = (10 \mu F)(14 V)=140 \mu C$

For the 2nd capacitor, we have

$Q_2 = C_2 V=(20 \mu F)(14 V)=280 \mu C$

After being charged, the two capacitors are disconnected from the battery, then they are connected in parallel, with the positive plate of C1 connected to the negative plate of C2. This means that charges will start to flow from one capacitor to the other (due to the opposite polarity), until the net charge on the plate of the capacitor with lower capacitance becomes zero.

Here, this means that $140 \mu C$ of charge will flow from C2 to C1: as a result, the final charge on the capacitor 1 will be

Q_1 = 0

While the final charge on capacitor 2 will be

$Q_2 = 280 - 140 = 140 \mu C$

In order to find the potential difference across each capacitor, we use the equation

$V=\frac{Q}{C}$

For capacitor 1, the charge is zero, so

V_1 = 0

For capacitor 2, the charge is

$Q_2 = 140 \mu C$

The capacitance is

$C=20 \mu F$

So, the potential difference across it will be

$V_2 = \frac{Q_2}{C_2}=\frac{140}{20}=7 V$

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

Answer from: Quest

Gravity bc it’s gravity

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Capacitors C1=10μF and C2=20μF are each charged to 14 V , then disconnected from the battery without...

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