Lompat ke konten Lompat ke sidebar Lompat ke footer
Beranda / what is the electric potential v at the origin

what is the electric potential v at the origin

Posting Komentar

Electric Potential

search icon

Practice

practice problem 1

The setup

A charge of −1.0 μC is located happening the y-axis 1.0 m from the origin at the coordinates (0,1) while a second charge of +1.0 μC is located on the x-Axis 1.0 m from the pedigree at the coordinates (1,0). Determine the value of the pursual quantities at the origin…

  1. the magnitude of the electric field
  2. the direction of the electric field
  3. the potential difference (assuming the potential is zero at infinite distance)
  4. the energy needed to bring a +1.0 μC charge to this position from infinitely far away

solution

  1. Since the charges are identical in order of magnitude and as far from the descent, we can ut ane computation for some charges.

    E = (9.0  × 109 N m2/C2)(1.0  × 10−6 C)
    (1.0 m)2

    Electric field lines step forward of positive charges and move in negative charges. At the origin, this results in an electric domain that points "leftover" (away from the positive exchange) and "up" (toward the negative charge). These two vectors imprint the legs of a 45°–45°–90° triangle whose sides are in the ratio 1:1:√2.

    ∑E =√2 × 9,000 N/C =12,700 N/C

  2. The solution

    Moving "awake" and to the "nigh" in equal amounts results in a 135° standard angle.

  3. Once over again, since the charges are identical in magnitude and as far from the origin, we lone need to compute cardinal count.

    V = (9.0  × 109 N m2/C2)(1.0  × 10−6 C)
    (1.0 m)

    Tense potential is a scalar amount. It doesn't have direction, but information technology does have sign of the zodiac. The positive charge contributes a positive latent and the negative charge contributes a negative potential difference. Add them up and vigil them cancel.

    ∑V =9,000 V − 9,000 V =0 V

  4. The potential difference at a point in blank is defined as the influence per unit charge required to move a screen electric charge to that location from infinitely far away.

    Algebra shows that work is charge times possible remainder. Since the potential at the origin is zero, no exercise is necessary to move a burden to this point.

    ∆UE  =q∆V
    ∆UE  =(1.0 × 10−6 C)(0 V)
    ∆UE  =0 J

practice problem 2

A proton (mass m, charge +e ) and an of import particle (great deal 4m , bear down +2e ) approach one another with the same initial speed v from an initially large distance. How close will these cardinal particles get to one another in front turning around?

solution

The kinetic energy of the moving particles is completely transformed into electric electric potential energy at the point of closest approach.

Ue  = K
k(e)(2e)  = 1  (m)v 2 + 1  (4m)v 2
r 2 2

Finish the algebra.

practice trouble 3

resume-v.pdf
The diagram below shows the location and charge of four isotropic small spheres. Recover the potential difference at the five points indicated with open circles. Use these results and symmetry to retrieve the potential at as many points as accomplishable without extra calculation. Write your results connected or near the points. Sketch at least 4 equipotential lines. Pick round values seperated away a regular interval. At least uncomparable of the lines should live disconnected.

The setup

answer

Employment the equation for the electric potential from a set of point charges.

Since each charge is the synoptical size, we can factor it out.

Systematic to salve screen realty, let's compute the merchandise of the constants once…

kq =(9 × 109 N m2/C2)(1 × 10−6 C) = (9,000 N m2/C)

and the sum of the distances to the four charges five times…

∑ 1  = ⎛
⎜
⎝		 1  + 1  + 1  + 1 ⎞
⎟
⎠
r 1 √8 m √8 m √8 m √8 m
∑ 1  = 1.41421…m−1
r 1
∑ 1  = ⎛
⎜
⎝		 1  + 1  + 1  + 1 ⎞
⎟
⎠
r 2 √8 m √8 m √40 m √40 m
∑ 1  = 1.02333…m−1
r 2
∑ 1  = ⎛
⎜
⎝		 1  + 1  + 1  + 1 ⎞
⎟
⎠
r 3 √20 m √20 m √68 m √68 m
∑ 1  = 0.68974…m−1
r 3
∑ 1  = ⎛
⎜
⎝		 1  + 1  + 1  + 1 ⎞
⎟
⎠
r 4 2 m 2 m √20 m √20 m
∑ 1  = 1.44721…m−1
r 4
∑ 1  = ⎛
⎜
⎝		 1  + 1  + 1  + 1 ⎞
⎟
⎠
r 5 2 m √20 m 6 m √52 m
∑ 1  = 1.02894…m−1
r 5

Give it at to each one of the five locations, summing up the contributions of the four degree charges.

V 1  = (9,000 N m2/C)(1.41421…m−1)
V 1  = 12,700 V
V 2  = (9,000 N m2/C)(1.02333…m−1)
V 2  = 9,210 V
V 3  = (9,000 N m2/C)(0.68974…m−1)
V 3  = 6,210 V
V 4  = (9,000 N m2/C)(1.44721…m−1)
V 4  = 13,000 V
V 5  = (9,000 N m2/C)(1.23605…m−1)
V 5  = 9,260 V

Memorialise the numbers at as many symmetric locations arsenic possible.

All the numbers in their correct locations

Sketch in the equipotentials.

The completed drawing

practice problem 4

Fission is the splitting of a with child atomic nucleus into cardinal roughly quits halves accompanied away the release of a large amount of money of energy. An minute core group can be modeled as a sphere whose charge is distributed uniformly across its whole bulk. Determine the vim discharged when a heavy nucleus undergoes nuclear nuclear fission using electrostatic principles.

  1. Derive an par for the electrostatic vitality needed to assemble a emotional orbit from an infinite swarm of infinitesimal charges located infinitely far away. (In else words, use calculus.) Allow R beryllium the sphere's radius, Q be its total charge, V be its volume, and ρ be its charge up density.
  2. Explicit the add together energy of two half-sized spheres in price of the vigor of one unhurt sphere. One-half-sized spheres have half the volume and half the institutionalize of a whole sphere (because electric charge density is assumed to be constant).
  3. Depend the energy free when a nucleus of uranium 235 (the isotope responsible for powering some nuclear reactors and cell organ weapons) splits into cardinal identical daughter nuclei. Kick in your final answer in the preferred unit for nuclear reactions, the megaelectronvolt. (A nucleus of 235 92 U has a wheel spoke of 5.8337 fm.)

solution

  1. One way to make a titanic sphere to add layers to an already alive smaller area. Calculus allows the States to start with an initial orbit with zero radius ( r 0 = 0), minimal brain damage layers thereto of little thickness (dr), and end up with a sphere with nonzero radius ( r =R ) by repeating the process an infinite number of times (∫). This calculus thing is pretty amazing.

    The electrostatic potential energy of two item charges is given by…

    where…

    U = electric potential energy
    k = the electricity constant
    q 1 = one point in time charge
    q 1 = some other point charge
    r = the separation between charges

    In our sphere of influence built up layer by layer, the first charge is a solid sphere with undifferentiated charge density.

    The second charge is a thin spherical blast with the same charge density.

    q 2 = ρ(4πr 2 dr)

    These two charges are in effect divided past the radius of the solid sphere. The energy equation then becomes a mess…

    R
    ⌠
    ⌡
    0
    U = k (ρ 4  πr 3) ρ(4πr 2 dr) 1
    3 r

    begging to represent simplified…

    R
    ⌠
    ⌡
    0
    U = 16π2ρ2 k r 4 dr
    3

    and solved.

    We should straightaway replace charge density with a more useful expression.

    ρ = Q  = Q  = 3Q
    V 4 3 πR 3 4πR 3

    In reality, it's the square of direction density we should eliminate.

    Indeed let's do it.

    U = 16π2 kR 5 9Q 2
    15 16π2 R 6

    Yay algebra!

  2. A half sized vault of heaven has incomplete the charge and half the mass merely non half the radius. It's the cube root of a half the radius.

     ⇒
    V  = 4  π ⎛
    ⎜
    ⎝    		 r ⎞3
    ⎟
    ⎠
    2 3 ∛2

    A uncomplete sphere then has energy equal to…

    U ½ = 3k(Q/2)2  = ∛2 U
    5(R/∛2) 4

    and two of them have energy up to…

    I think information technology's to a greater extent interesting to express the unusual fraction as a denary.

    2U ½  = ∛2  = 0.62996…
    U 2

    Splitting a charged sphere in half reduces potential energy to 63% — or results in a loss of 37%, if you prefer. (This assumes the cardinal spheres are infinitely far away from each past, so their fundamental interaction adds no additional prospective energy.)

  3. Hera's how I'd like to approach this problem. Start by determining the electric potential energy of a 235 92U core exploitation the equation derived in part a.

    U = 3(8.99 × 109 N m2/C2)(92 × 1.60 × 10−19 C)2
    5(5.8337 ×10−15 m)

    Convert that into megaelectronvolts by dividing by the unproblematic charge (to get it into electronvolts) and as wel by a million (since the prefix mega means a million).

    U = 2.00 × 10−10 J
    (1.60 × 10−19 C)(1,000,000)

    Then take 37% of that.

    ∆U =0.370(1252 MeV) = 463 MeV

    This is about twice what I expected, but I'll have to figure out the discrepancy some opposite time.

No condition is perm.

what is the electric potential v at the origin

Source: https://physics.info/electric-potential/practice.shtml#:~:text=Since%20the%20potential%20at%20the,a%20charge%20to%20this%20point.

Posting Komentar untuk "what is the electric potential v at the origin"