answer.
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
slamgirl [31]
1 year ago
5

A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.Ex

press your answer in terms of the given quantities and appropriate constants.1. What is the electric field at the surface of the sphere?E=2.What is the electric potential at the surface of the sphere?V=
Physics
1 answer:
amid [387]1 year ago
3 0

Answer:

E =   k*Q₁/R₁² V/m

V =  k*Q₁/R₁ Volt

Explanation:

Given:

- Charge distributed on the sphere is Q₁

- The radius of sphere is R₁

- The electric potential at infinity is 0

Find:

What is the electric field at the surface of the sphere?E.

What is the electric potential at the surface of the sphere?V

Solution:

- The 3 dimensional space around a charge(source) in which its effects is felt is known in the electric field.

- The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.  

- If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

                                        F = k*Q₁/R₁²

- Then the electric field at that point is

                                        E =  F/1

                                        E =  k*Q₁/R₁²  V/m

- The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

- Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

                                        V =  k*Q₁/R₁  Volt

You might be interested in
Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu
Anit [1.1K]
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>

m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... y direction

Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 =  625N^2 =>

Net force = √625 = 25N

F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2

Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. </span>

<span>m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N


Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2

Answer: 6.08 m/s^2


</span>
8 0
1 year ago
Read 2 more answers
Light from a distant star is collected by a concave mirror that has a radius of curvature of 150 cm. How far from the mirror is
andrey2020 [161]
The focal point of a mirror is half of the radius of curvature. We can use the formula,                                                                                                                   1/f = 1/v + 1/u                                                                                                          where f is the focal length , v is the image distance and u is object distance        The distance of the star is assumed to be incredibly far away and 1 divided by a really big number is approximately zero, thus;                                                     1/f = 1/v  = 1/75 = 1/v                                                                                             therefore; the image is formed 75 cm infront of the mirror
4 0
2 years ago
Under the Big Top elephant, Ella (2500 kg), is attracted to Phant, the 3,000 kg
Vladimir [108]

Under the Big Top elephant, Ella (2500 kg), is attracted to Phant, the 3,000 kg elephant. They are separated by 8

4 0
1 year ago
A cup slides off a 1.1\,\text m1.1m1, point, 1, start text, m, end text high table with a speed of 1.3\,\dfrac{\text m}{\text s}
Tju [1.3M]

Answer:

<h3>0.145m</h3>

Explanation:

Using the equation of motion formula y = ut + 1/2gt² where;

y is the horizontal displacement

u is the initial velocity

g is the acceleration due to gravity

t is the time

To calculate the horizontal displacement, we need to first get the time t. Using the equation:

v = u+gt

1.3² = 0+(9.8)t

1.69 = 9.8t

t = 1.69/9.8

t = 0.172s

Substituting the time into the equation above to get y

y = 0+1/2(9.8)(0.172)²

y = 4.905(0.029584)

y = 0.145m

Hence the cup's horizontal displacement during the fall is 0.145m

8 0
1 year ago
A fire engine is moving south at 35 m/s while blowing its siren at a frequency of 400 Hz.
vodomira [7]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as

f_d= f_s \frac{(v+v_d)}{(v-v_s)}

Here,

f_d=frequency received by detector

f_s=frequency of wave emitted by source

v_d=velocity of detector

v_s=velocity of source

v=velocity of sound wave

Replacing we have that,

f_d = 400(\frac{(343+18)}{(343-35)})

f_d=422 Hz

Therefore the frequencty that will hear the passengers is 422Hz

8 0
2 years ago
Other questions:
  • Engineers who design battery-operated devices such as cell phones and MP3 players try to make them as efficient as possible. An
    9·1 answer
  • If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol
    13·1 answer
  • The posted speed limit on the road heading from your house to school is45 mi/h, which is about 20 m/s. If you live 8 km (8,000 m
    15·2 answers
  • A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w
    8·1 answer
  • While a car is stopped at a traffic light in a storm, raindrops strike the roof of the car. The area of the roof is 5.0 m2. Each
    13·1 answer
  • A solenoid of length 0.700m having a circular cross-section of radius 5.00cm stores 6.00 μJ of energy when a 0.400-A current run
    10·2 answers
  • Snowboarder Jump—Energy and Momentum
    9·1 answer
  • A closely wound rectangular coil of 80 turns has dimensions 25.0 \rm cm by 40.0 \rm cm. The plane of the coil is rotated from a
    9·2 answers
  • In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Suppose that during 115 m of her walk Salvador is observed to
    6·1 answer
  • the container is filled with liquid. the depth of liquid is 60 cm. if it is exerting the pressure of 2000pa. calculate the densi
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!