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sattari [20]
1 year ago
7

The voltage between two parallel plates separated by a distance of 3. 0 cm is 120 v. The electric field between the plates is?

Physics
1 answer:
Lapatulllka [165]1 year ago
6 0

The electric field between plates is 4000V/m.

An electric field (sometimes E-field) is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles.

The value of the electric field has dimensions of force per unit charge. In the metre-kilogram-second and SI systems, the appropriate units are newtons per coulomb, equivalent to volts per metre.

The voltage between points A and B is

V=E.d

E =V/d  (uniform E- field only)  

where  d is the distance from A to B, or the distance between the plates.

Given:

distance d = 3 cm

voltage V = 120 V

Electric field E = V/d

                     E = 120 V / 3cm

                     E = 40 V / 1 cm               [ 1 cm = 1/100 m ]

                    E = 4000 V/m.

Learn more about Electric field here:

brainly.com/question/8971780

#SPJ4

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Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixe
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Answer:

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}\ J.

Explanation:

Given charges are:

\rm q_1 = +6e.\\q_2 = -6e.

The electric potential energy of a charge due to the electric field of another charge is given by

\rm EPE=\dfrac{kq_1q_2}{r}.

where,

  • k = Coulomb's constant, having value = \rm 9\times 10^9\ Nm^2/C^2.
  • r = distance between the charges.

When the charges are infinite distance apart, \rm r = \infty,

\rm EPE_{initial} = \dfrac{kq_1q_2}{r}=0\ J.

When the charges are \rm 5.61\times 10^{-12}\ m apart, \rm r=5.61\times 10^{-12}\ m,

\rm EPE_{final}=\dfrac{kq_1q_2}{r}\\=\dfrac{(9\times 10^9)\times (+6e)\times (-6e)}{5.61\times 10^{-12}}\\=-5.775\ e^2\times 10^{22}.

Here, e is the charge on one electron, such that, \rm e = -1.6\times 10^{-19}\ C.

Therefore,

\rm EPE_{final}=-5.775\times (-1.6\times 10^{-19})^2\times 10^{22} = -1.478\times 10^{-15}\ J.

Thus,

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}-0=-1.478\times 10^{-15}\ J.

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3 years ago
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