Vikolers6

2022-01-06

A 100 pF capacitor is charged to a potential difference of 50 V, and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the potential difference across the first capacitor drops to 35 V, what is the capacitance of this second capacitor?

David Clayton

Beginner2022-01-07Added 36 answers

Initial charge on the C1:

${q}_{i}1={C}_{1}{V}_{i}=100\cdot 50=5000pC$

final charge:

${q}_{f}1={C}_{1}{V}_{f}=100\cdot 35=3500pC$

After connecting the capacitors, the charge on the second capacitor will be $q}_{2$:

The electric potentials are equal because the capacitors are connected in parallel:

${V}_{2}=35V$

the capacitance of the second capacitor is:

${C}_{2}=\frac{{q}_{2}}{{V}_{2}}=\frac{1500}{35}=43pC$

kaluitagf

Beginner2022-01-08Added 38 answers

The charge initially on the charged capacitor is given by $q={C}_{1},{V}_{0}$ , where ${C}_{1}=100pF$ is the capacitance and ${V}_{0}=50V$ is the initial potential difference. After the battery is disconnected and the second capacitor wired in parallel to the first, the charge on the first capacitor is ${q}_{1}={C}_{1}V$ , where $V=35V$ is the new potential difference. Since charge is conserved in the process, the charge on the second capacitor is $q}_{2}=q-{q}_{1$ , where $C}_{2$ is the capacitance of the second capacitor. Substituting $C}_{1}{V}_{0$ for q and ${C}_{1}V$ for $q}_{1$ , we obtain ${q}_{2}={C}_{1}({V}_{0}-V)$ . The potential difference across the second capacitor is also V, so the capacitance is:

${C}_{2}=\frac{{q}_{2}}{V}=\frac{{V}_{0}-V}{V}{C}_{1}=\frac{50V-35V}{35V}\left(100pF\right)=43pF$

karton

Expert2022-01-10Added 613 answers

Consider the initial charge on the capacitor $(q)={C}_{1}{V}_{0}=(100PF)(50V)=5000pC$

The voltage on the first capacitor is now ${q}_{1}={C}_{1}V$ when the battery is detached and the second capacitor wire is parallel to the first.

The potential difference across the first capacitor drops to $(V)=35V,$

The $n{q}_{1}=(100pF)(35V)=3500pC$

Since the charge is conserved in this process, then thecharge on the second capacitor is ${q}_{2}=q-{q}_{1}=5000pC-3500pC=1500pC$. The capcacitance of the second capacitor is ${C}_{2}={q}_{2}/V=1500pC/35V=42.85pF\approx 43pF$

22+64

When a cold drink is taken from a refrigerator, its temperature is 5 degree C. After 25 minutes in a 20 degree C room its temperature has increased to 10 degree C. What is the temperature of the drink after 50 minutes?

How many minutes are there in $2\frac{3}{5}$ hours?

Find the answer for $64$ power $\frac{1}{6}$.

The following law describes the relationship between gas volume and pressure: ________. A)Boyle's law; B)Henry's law; C)Charles' law; D)Dalton's law

If an electric switch is open then no current flows through the circuit.

A)True;

B)FalseThe price elasticity of supply of a good is 0.8. Its price rises by 50 per cent. Calculate the percentage increase in its supply.

How many quarts are in 10 gallons?

When the net force acting on an object is zero, the total momentum of the system will:

A)remain constant;

B)increase with time;

C)decrease with time;

D)None of theseHow many centimeters are in four meters?

A notebook computer has a mass of 2.25 kilograms. About how many pounds does the notebook weigh?

If the dot product of two non-zero vectors is zero, then the vectors A)are parallel to each other. B)are perpendicular to each other. C)can have any orientation. D)are anti-parallel to each other.

Which is longer 5 miles or 10 kilometers?

9 grams is how many milligrams?

How to write $673.5$ in scientific notation?