1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Jet001 [13]
3 years ago
7

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

Physics
2 answers:
Len [333]3 years ago
8 0

Answer:

108.66V

Explanation:

E = 1.20*10⁻³J

q = 6.70μC = 6.7*10⁻⁶C

V =?

K.E = 4.72 * 10⁻⁴J

E = q∇v

but there's difference in energy in moving the electron from point A to B. The electron has an initial energy of 4.72*10⁻⁴J.

E = Q - KE

E = 1.20*10⁻³ - 4.72*10⁻⁴ = 7.28*10⁻⁴J

E = q∇v

∇v = E / q

∇v = 7.28*10⁻⁴ / 6.7*10⁻⁶

∇v = 108.66V

the change in potential difference from point A to B was 108.66V

ASHA 777 [7]3 years ago
4 0

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

In this problem, we have

W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

Solving the formula for \Delta V, we find

\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

You might be interested in
A sphere of radius R = 0.295 m and uniform charge density -151 nC/m^3 lies at the center of a spherical, conducting shell of inn
cupoosta [38]

Answer:

a) -1.27*10³ N/C b) 0 c) -0.21*10³ N/C d) 0.1*10³ N/C

Explanation:

a) r = 0.76R

As this distance is inside the sphere, we need to know how much charge is enclosed within this distance for the center, as follows:

Q = ρ*V(r) = ρ*\frac{4}{3} *\pi *r^{3}

where r = 0.760* R = 0.760* 0.295 m = 0.224 m, and ρ = -151 nC/m³

Q = -151e-9 *\frac{4}{3} *\pi *0.224m^{3} = -7.11e-9 C

Applying Gauss' Law to a spherical gaussian surface of r= 0.76R, as the electric field is radial, and directed inward, we can write the following equation:

E*A = Q/ε₀, where Q= -7.11 nC, A= 4*π*(0.76R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{-7.11e-9C}{(0.76*0.295m)^{2}} =-1.27e3 N/C

⇒ E = -1.27*10³ N/C

b) r= 3.90 R

As this distance falls inside the conducting shell, and no electric field can exist within a conductor in electrostatic condition, E=0

c) r = 2.8 R

As this distance falls between the sphere and the inner radius of the shell, we can calculate the electric field, applying Gauss' law to a gaussian surface of radius equal to r= 2.80 R.

First we need to find the total charge of the sphere, as follows:

Q = ρ*V =

Q = -151e-9 *\frac{4}{3} *\pi *0.295m^{3} = -16.2e-9 C

In the same way that for a) we can write the following expression:

E*A = Q/ε₀, where Q= -16.2 nC, A= 4*π*(2.8R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{-16.2e-9C}{(2.8*0.295m)^{2}} =-0.21e3 N/C

⇒ E = -0.21*10³ N/C

d) r= 7.30 R

In order to find the electric field at this distance, which falls beyond the outer radius of the shell, we need to find the total charge on the outer surface.

As the sphere has a charge of -16.2 nC, and the total charge of the conducting shell is 66.7nC, in order to make E=0 inside the shell, the total charge enclosed by a gaussian surface with a radius larger than the inner radius of the shell and shorter than the outer one, must be zero, which means that a charge of +16.2 nC must be distributed on the inner surface of the shell.

This leaves an excess charge on the outer surface of the shell as follows:

Qsh = 66.7 nC - 16.2 nC = 50.5 nC

Now, we can repeat the same process than for a) and c) as follows:

E*A = Q/ε₀, where Q= 50.5 nC, A= 4*π*(7.3R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{50.5e-9C}{(7.3*0.295m)^{2}} =0.1e3 N/C

⇒ E = 0.1*10⁻³ N/C

6 0
3 years ago
On a stormy night, a lighthouse emits a light that must travel through air, rain, and fog. As the light travels through these di
vlada-n [284]

Answer:

HOOOOOOOYAAAAHHHHH

Explanation:

5 0
3 years ago
Read 2 more answers
Why do astronomers find it difficult to detect individual exoplanets?
Maurinko [17]
Light from the stars, because the orbits make it difficult to see them. 
6 0
3 years ago
Read 2 more answers
Which of the following statements best describes the second law of
Serga [27]

Answer:

B

Explanation:

Heat flows from hot to cold to lower the temperature of hot areas and increase temperature of cold areas. The end result is that the 2 areas have the same temperature, thus increasing entropy.

7 0
3 years ago
A student took a calibrated 250.0 gram mass, weighed it on a laboratory balance, and found it read 266.5 g. What was the student
Kruka [31]

Answer:

B. 6.6%

Explanation:

The percentage error of a measurement can be calculated using the formula;

Percent error = (experimental value - accepted value / accepted value) × 100

In this question, the calibrated 250.0 gram mass is the accepted value while the weighed mass of 266.5 g is the experimental or measured value.

Hence, the percentage error can be calculated thus;

Percent error = (266.5-250.0/250.0) × 100

Percent error = 16.5/250 × 100

Percent error = 0.066 × 100

Percent error = 6.6%

7 0
3 years ago
Other questions:
  • A student decides they need a cup of coffee. The energy lost as the coffee cools down is _____ the energy gained by the surround
    5·2 answers
  • A train travels at a speed of 30 miles/hour and traveled a distance of 240 miles. How
    14·1 answer
  • UVC light used in sterilizers, has wavelengths between 100 to 280 nm. If a certain UVC wave has a wavelength of 142.9 nm, what i
    13·1 answer
  • What is the hardy-weinberg principle?
    12·1 answer
  • A ball is rolling a 5 m/s down a hill the ball has 20 J of energy what is the mass of the ball
    8·1 answer
  • A transformation of ΔSTV results in ΔUTV. Which transformation maps the pre-image to the image?
    13·2 answers
  • Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.8 km/h due east. Runner B is initia
    12·1 answer
  • Select the correct answer. What happens to the molecules within a gas when the gas condenses?
    7·1 answer
  • A ray of light travelling obliquely from a rarer medium to a denser medium goes _______​
    8·1 answer
  • Do two bodies have to be in physical contact to exert a force upon one another?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!