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Yuliya22 [10]
2 years ago
7

By standard convention, both the electric potential and the the electric potential energy between two charges is taken to be zer

o in what configuration?
Physics
1 answer:
shusha [124]2 years ago
6 0

Answer: at when distance r = infinity.

Explanation: The formulae for the electric potential of an electric charge to an arbitrary point is given by the formulae below

V = q/4πεr

V = electric potential (volts)

q = magnitude of electric charge

ε = permittivity of free space

r = distance between arbitrary point and charge.

In the equation above, it can be seen that only electric potential (v) and distance (r) is a variable, and there is an inverse relationship between them (an increase in one leads to a decrease in the other)

Thus to have zero value of electric potential (v= 0) we have to have the largest value of r ( r = infinity).

Same goes for electric potential energy between two charges, the formulae is given below as

W = q1 *q2/4πεr

W= electric potential energy

q1 = magnitude of first charge.

q2 = magnitude of second charge

ε = permittivity of free space

r = distance between arbitrary point and charge.

Also, all values are constant aside from electric potential energy (w) and distance (r) which have an inverse relationship.

Thus to have zero value of electric potential energy (w =0), we have to get an infinite value of distance ( r =infinity)

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A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
adell [148]

Answer:

r = 0.114 m

Explanation:

To find the speed of the proton, from conservation of energy, we know that

KE = PE

Thus, we have;

(1/2)mv² = qV

Where;

V is potential difference = 1kv = 1000V

q is charge on proton which has a value of 1.6 x 10^(-19) C

m is mass of proton with a constant value of 1.67 x 10^(-27) kg

Let's make the velocity v the subject;

v =√(2qV/m)

v = √(2•1.6 x 10^(-19)•1000)/(1.67 x 10^(-27))

v = 4.377 x 10^(5) m/s

Now to calculate the radius of the circular motion of charge we know that;

F = mv²/r = qvB

Thus, mv²/r = qvB

Divide both sides by v;

mv/r = qB

Thus, r = mv/qB

Value of B from question is 0.04T

Thus,

r = (1.67 x 10^(-27) x 4.377 x 10^(5))/(1.6 x 10^(-19) x 0.04)

r = 0.114 m

r = 8.76 m

8 0
1 year ago
At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is inc
lys-0071 [83]

Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

I² = 10.2

I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

5 0
2 years ago
A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc
Harman [31]

Answer:

E=88200\ J

Explanation:

Given:

  • mass of car, m=1500\ kg
  • distance of skidding after the application of brakes, d=15\ m
  • coefficient of kinetic friction, \mu_k=0.4

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

f=\mu_k.N

where N = normal reaction by ground on the car

f=0.4\ties 1500\times 9.8

f=5880\ N

<em>Now from the work-energy equivalence:</em>

E=f.d

E=5880\times 15

E=88200\ J is the dissipated energy.

3 0
2 years ago
when you drop a pebble from height h, it reaches the ground with kinetic energy k if there is no air resistance. from what heigh
marysya [2.9K]

Answer:

From the initial height h

Explanation:

When a material or substance is drop from a height h, it possesses potential energy, immediately it is dropped from that height, the potential energy is gradually converted to kinetic energy, it gets to a point where the potential energy equals the kinetic energy, as the material touches the ground, all potential energy has been converted to kinetic energy already

6 0
2 years ago
Onur drops a basketball from a height of 10\,\text{m}10m10, start text, m, end text on Mars, where the acceleration due to gravi
Doss [256]

Answer:

Explanation:

Given that,

Basket ball is drop from height

H=10m

It is dropped on planet mass

And the acceleration due to gravity on Mars is given as

g= 3.7m/s²

Time taken for the ball to reach the ground

Initial velocity of the body is zero

u=0m/s

Using equation of motion: free fall

H = ut + ½gt²

10 = 0•t + ½ × 3.7 ×t²

10 = 0 + 1.85t²

10 = 1.85t²

Then, t² =10/1.85

t² = 5.405

t = √ 5.405

t = 2.325seconds

So the time the ball spend on the air before reaching the ground is 2.325 seconds

5 0
2 years ago
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