Ask question

Ask question

All categories

2 years ago

15

Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite ea

ch other. Between the electrodes but not near their edges, the electric field strength is 2.5 × 106 N/C. What is the magnitude of the charge on each electrode? (ε0 = 8.85 × 10-12 C2/N ∙ m2)

1 answer:

serious [3.7K]2 years ago

7 0

Answer:

1.77 x 10^-8 C

Explanation:

Let the surface charge density of each of the plate is σ.

A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2

d = 2 mm

E = 2.5 x 10^6 N/C

ε0 = 8.85 × 10-12 C2/N ∙ m2

Electric filed between the plates (two oppositively charged)

E = σ / ε0

σ = ε0 x E

σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2

The surface charge density of each plate is ± σ / 2

So, the surface charge density on each = ± 22.125 x 10^-6 / 2

= ± 11.0625 x 10^-6 C/m^2

Charge on each plate = Surface charge density on each plate x area of each plate

Charge on each plate = ± 11.0625 x 10^-6 x 16 x 10^-4 = ± 1.77 x 10^-8 C

You might be interested in

A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov

vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

$Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}$

$Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V$

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

$R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200$

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.

if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

$Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V$

$Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V$

$R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140$

$R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260$

so.

$V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}$

4 0

2 years ago

Suppose two astronauts on a spacewalk are floating motionless in space, 3.0 m apart. Astronaut B tosses a 15.0 kg IMAX camera to

marta [7]

Answer:

$\frac{ 112.5}{15+m_{A}}=v_{f}$

(we need the mass of the astronaut A)

Explanation:

We can solve this by using the conservation law of the linear momentum P. First we need to represent every mass as a particle. Also we can simplify this system of particles by considering only the astronaut A with an initial speed $v_{iA}$ of 0 m/s and a mass $m_{A}$ and the IMAX camera with an initial speed $v_{ic}$ of 7.5 m/s and a mass $m_{c}$ of 15.0 kg.

The law of conservation says that the linear momentum P (the sum of the products between all masses and its speeds) is constant in time. The equation for this is:

$P_{i}=p_{ic}+p_{iA}\\P_{i}=m_{c}v_{ic}+m_{A} v_{iA}\\P_{i}=15*7.5 + m_{A}*0\\P_{i}=112.5 \frac{kg.m}{s}$

By the law of conservation we know that $P_{i} =P_{f}$

For $P_{f}$ (final linear momentum) we need to treat the collision as a plastic one (the two particles stick together after the encounter).

So:

$P_{i} =P_{f}=112.5\\$

$112.5=(m_{c}+m_{A})v_{f}\\\frac{ 112.5}{m_{c}+m_{A}}=v_{f}\\\frac{ 112.5}{15+m_{A}}=v_{f}$

3 0

2 years ago

Th e heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its tem

Darina [25.2K]

Answer:

Q=1005 J

t= 0.67 sec

Explanation:

Lets take condition of room is 1 atm and 25°C.

Heat capacity ,c = 21 J /K.mol

If we assume that air is ideal gas that

P V = n R T

$V= 5.5\times 6.5\times 3\ m^3$

$V=107.25\ m^3$

$V=107.25\times 1000 L$

V= 107250 L

At STP number of moles given as

$n=\dfrac{V}{V_{at\ S.T.P.}}$

V=22.4 L at S.T.P.

$n=\dfrac{107250}{22.4}\ moles$

n=4787.94 moles

n= 4.784 Kmoles

So heat required to raise 10°C temperature

Q = n x c x ΔT

Q = 4.78794 x 21 x 10

Q=1004.64 J

Time t

t= Q/P

P= 1.5 KW

t = 1.004.64 /1.5

t= 0.66 sec

4 0

2 years ago

Read 2 more answers

Find the average power Pavg created by the force F in terms of the average speed vavg of the sled.

Katena32 [7]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created is $P_{avg} = F * v_{avg}$

Explanation:

From the question we are told that

The that the average power is mathematically represented as

$P_{avg} = \frac{W }{\Delta t }$

Where W is is the Workdone which is mathematically represented as

$W = F * s$

Where F is the applies force and s is the displacement due to the force

So

$P_{avg} = \frac{F *s }{\Delta t }$

Now this displacement can be represented mathematically as

$s = v_{avg} * \Delta t$

Where $v_{avg }$ is the average velocity and $\Delta t$ is the time taken

So

$P_{avg} = \frac{F *v_{avg} * \Delta t }{\Delta t }$

=> $P_{avg} = F * v_{avg}$

3 0

2 years ago

1. A car is 140 kg, and drove East 13.5 m/s. Car B is 157 kg and drove West at 10.9 m/s.

vaieri [72.5K]

The velocity would switch on the cars

4 0

2 years ago