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Citrus2011 [14]
3 years ago
11

An electric field exerts an electrostatic force of magnitude 1.5 x 10-14 newton on an electron within the field. what is the mag

nitude of the electric field strength at the location of the electron?
a.2.4 x 10-33 n/c
b.1.1 x 10-5 n/c
c.9.4 x 104 n/c
d.1.6 x 1016 n/c
Physics
1 answer:
stira [4]3 years ago
7 0
The electrostatic force on the electron is given by
F=qE
where q is the electron charge and E the intensity of the electric field.

So, in order to find E, we just need to re-arrange the formula and use F=1.5 \cdot 10^{-14} N:
E= \frac{F}{q} = \frac{1.5 \cdot 10^{-14}N}{1.6 \cdot 10^{-19}C}=9.4 \cdot 10^4 N/C

So, the correct answer is C.
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An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm. A 21
Inga [223]

Answer:

12353 V m⁻¹ = 12.4 kV m⁻¹

Explanation:  

Electric field between the plates of the parallel plate capacitor depends on the potential difference across the plates and their distance of separation.Potential difference across the plates V over the distance between the plates gives the electric field between the plates. Potential difference is the amount of work done per unit charge and is given here as 21 V. Electric field is the voltage over distance.

E = V ÷ d = 21 ÷ 0.0017 = 12353 V m⁻¹

8 0
3 years ago
An uncharged, nonconducting, hollow sphere of radius 10.0cm surrounds a 10.0-μC charge located at the origin of a Cartesian coor
MrMuchimi

The electric flux through the hole is 56.45\ webber .

  • Electric flux is the number of electric field lines cutting through the surface and is measured as surface intregal of electric field over that surface
  • Mathematically it is given by \phi_E=E.A \ Nm^2/C where E is the electric field and A is the area.
  • Gauss's law states that electric flux through closed surface is equal to the 1 / ε₀ times the charge enclosed by that surface which is given by  Ф = q / ε₀ where q is the central charge and ε₀ is the permittivity of the medium.

It is given , hollow sphere of radius 10.0cm surrounds a 10.0-μC charge.

The whole surface of hollow sphere = 4\pi r^2

                                                            = 4\times 3.14\times  (10 \times  10^{-2})^2 \\\\= 12.56\times 10^{-2} m^2

Area of the hole ( both side ) = 2\times \pi  r^2

                                               = 2\times 3.14 \times  (10^-^3)^2\\= 6.28 \times 10^-^6 m^2

According to Gauss's theorem, the flow from a particular charge in the center is given by

 \phi=  \frac{10\times10^-^6}{8.85\times 10^-^1^2}\\\\\phi=1.13\times10^6

This flux flows through the surface of the sphere, so the flux  per unit area which is given by

= \frac{ 1.13\times 10^6 }{ 12.56\times 10^-^2} \\\\= 8.99 \times 10^6 \  weber / m^2

Flux through area of hole is given by :

=  8.99\times10^6 \times6.28 \times 10^-^6\\ = 56.45 \ weber

Learn about more electric flux here :

brainly.com/question/26289097

#SPJ4

8 0
1 year ago
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

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Explanation:

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Answer:

I think your bones, muscles, and joints your welcome :)

Explanation:

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