Ask question

Ask question

All categories

3 years ago

12

A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positiv

e plate of each connected to the negative plate of the other. What is the final charge on the 6.0-mF capacitor? Group of answer choices

2 answers:

arsen [322]3 years ago

7 0

Answer:

0.192 C

Explanation:

The charge in a capacitor is given as,

Q = CtVt..................... Equation 1

Where Q = Charge, Ct = Effective Capacitance of the capacitors, Vt = Effective Voltage.

Note: If the capacitors are connected to each other with the positive plate of each connected to the negative plate of the other means that the capacitors are connected in series

The combined capacitor in series is given as,

1/Ct = 1/C1 + 1/C2

Where C1 = Capacitance of the first capacitor, C2 = Capacitance of the second capacitor.

Ct = C1C2/(C1+C2)...................... Equation 2

Given: C1 = 4.0 mF, C2 = 6.0 mF

Substitute into equation 2

Ct = (4×6)/(4+6)

Ct = 24/10

Ct = 2.4 mF.

Also,

Vt = V1 + V2................... Equation 4

Where V1 = Voltage in the first capacitor, V2 = Voltage in the second capacitor.

Given: V1 = 50 V, V2 = 30 V

Vt = 50+30

Vt = 80 V.

Substitute the value of Vt and Ct into equation 1

Q = 80(2.4)

Q = 192 mC

Q = 0.192 C.

Since both capacitors are in series, The same quantity of charge flows through them.

Hence the final charge on the 6.0 mF capacitor = 0.192 C

Juli2301 [7.4K]3 years ago

5 0

Answer:

<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>

Explanation:

The initial charge on 4 mF capacitor = 4 mf x 50 V = 200 mC

The initial Charge on 6 mF capacitor = 6 mf x 30 V =180 mC

Since the negative ends are joined together the total charge on both capacity would be;

q = $q_{1} -q_{2}$

q = 200 - 180

q = 20 mC

In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage

q = (4 x V) + (6 x V)

20 = 10 V

V = 2 V

For the final charge on 6.0 mF;

q = CV

q = 6.0 mF x 2 V

q = 12 mC

Therefore the final charge on the 6.0 mF capacitor would be 12 mC

You might be interested in

Imagine that an electron in an excited state in a nitrogen molecule decays to its ground state, emitting a photon with a frequen

mash [69]

Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.

The energy of the emitted photon is given by:
$E=hf$
where
h is the Planck constant
f is the photon frequency
Substituting $f=8.88 \cdot 10^{14}Hz$ , we find
$E=hf=(6.6 \cdot 10^{-34} Js)(8.88 \cdot 10^{14} Hz)=5.86 \cdot 10^{-19} J$

This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
$\Delta E= -5.86 \cdot 10^{-19} J$

6 0

3 years ago

Read 2 more answers

Since the coefficient of friction is less than 1, what does that mean about the normal force and the force of friction?

ludmilkaskok [199]

Answer:

there is friction between the two things

Explanation:

7 0

2 years ago

The sport with the fastest moving ball is jai alai, where measured speeds can be 286 km/h. If a professional jai alai player fac

cricket20 [7]

Answer:

d= 794.4 cmExplanation:

Given that

Speed ,V= 286 km/h

$=286\times \dfrac{1000}{3600}\ m/s$

V=79.44 m/s

Given that time ,t= 100 ms

t= 0.1 s

We know that ( if acceleration is zero)

Distance = Speed x time

d= V t

Now by putting the values in the above equation

d = 79.44 x 0.1 m

d= 7.944 m

We know that 1 m = 100 cm

d= 794.4 cm

5 0

3 years ago

Mass of trolley (m):

Nookie1986 [14]

Answer:

ij

Explanation:

4 0

3 years ago

A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .

Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}$

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
$B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}$

Taking out constants from the integral:
$B =\frac{\mu_0 i}{4\pi R^2} \int ds$

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

$B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds$

Evaluate:
$B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}$

Plugging in our givens to solve for the magnetic field strength of one loop:

$B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T$

Multiply by the number of loops to find the total magnetic field:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

b)

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}$

Using the diagram, if 'z' is the point's height from the center:

$r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}$

Substituting this into our expression:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }$

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds$

Evaluate:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Multiplying by the number of loops:
$B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Plug in the given values:
$B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}$

5 0

2 years ago

Read 2 more answers