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loris [4]
4 years ago
5

PLLLZ HELP Compare and contrast electric potential energy and electric potential difference? Please use 2 content related senten

ces
Physics
1 answer:
masha68 [24]4 years ago
5 0

The electric potential difference is the electric potential energy per unit charge

Explanation:

First of all, we define the concept of electric potential. The electric potential is a measure of the gradient of the electric field at a certain point of the space. The electric potential at a distance r from a positive charge of magnitude q is given by

V(r) = \frac{kq}{r}

where k is the Coulomb's constant.

Now we can define the electric potential energy and the electric potential difference:

  • Electric potential energy is the energy possessed by a charge due to the presence of an electric field. For a charge of magnitude q immersed in an electric field, its potential energy is given by U=qV, where V is the electric potential at the location of the charge.
  • The electric potential difference is simply the difference in electric potential between two points in the space. For instance, if the potential at point A is V(A) and the potential at point B is V(B), then the potential difference is \Delta V = V(A)-V(B)

The electric potential energy is also defined as the work done on a charge q moved through a potential difference of \Delta V. Consequently, the potential difference \Delta V represents the work per unit charge done, i.e. the work done when moving a unitary charge through a potential difference \Delta V.

Learn more about potential difference and current:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

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MAXImum [283]

Answer:

The final speed of the electron = 2.095×10⁸ m/s

Explanation:

From newton's fundamental equation of dynamics,

F = ma ........................Equation 1

Where F = force, m = mass of the electron, a = acceleration of the electron.

making a the subject of the equation,

a = F/m.................... Equation 2

Given: F = 4.0×10⁻¹⁵ N,

Constant: m =  9.109×10⁻³¹ kg.

Substituting into equation 2

a = 4.0×10⁻¹⁵/9.109×10⁻³¹

a = 4.39×10¹⁶ m/s².

Using newton's equation of motion,

v² = u²+2as .......................... Equation 3

Where v = final velocity of the electron, u = initial velocity of the electron, a = acceleration of the electron, s = distance covered by the electron.

Given: u = 0 m/s(at rest), s = 50 cm = 0.5 m, a = 4.39×10¹⁶ m/s²

Substituting into equation 3

v² = 0² + 2(0.5)(4.39×10¹⁶)

v = √(4.39×10¹⁶)

v = 2.095×10⁸ m/s

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7 0
3 years ago
At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field t
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Answer:

The frequency of the coil is 7.07 Hz

Explanation:

Given;

number of turns of the coil, 200 turn

cross sectional area of the coil, A = 300 cm² = 0.03 m²

magnitude of the magnetic field, B = 30 mT = 0.03 T

Maximum value of the induced emf, E = 8 V

The maximum induced emf in the coil is given by;

E = NBAω

Where;

ω is angular frequency = 2πf

E = NBA(2πf)

f = E / 2πNBA

f = (8) / (2π x 200 x 0.03 x 0.03)

f = 7.07 Hz

Therefore, the frequency of the coil is 7.07 Hz

7 0
4 years ago
In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e
Zanzabum

Answer:

The  velocity is  v_t  =  0.02175 \  m/s

Explanation:

From the question we are told that

   The  mass of the bullet is  m_b  =  0.024 \  kg

    The initial speed of the bullet is  u_b  =  1200 \  m/s

   The mass of the target is  m_t  =  320 \  kg

    The  initial velocity of target is  u_t  =  0  \ m/s

    The  final velocity of the bullet is  is  v_b  =  910 \  m/s

   

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=>    v_t  =  0.02175 \  m/s

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