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creativ13 [48]
1 year ago
12

Q/C S Starting with the definition of work, prove that at every point on an equipotential surface, the surface must be perpendic

ular to the electric field there.
Physics
1 answer:
AveGali [126]1 year ago
7 0

It is proved that at every point on an equipotential surface, the surface must be perpendicular to the electric field.

<h3>What is meant by equipotential surface?</h3>
  • An equipotential surface is any surface where the potential is constant. In other words, any two points on an equipotential surface have the same potential difference.
  • Equipotential surfaces have the following important properties: 1. The work done in moving a charge across an equipotential surface is equal to zero.
  • An equipotential surface is one on which all of the points on it have the same electric potential.
  • This means that a charge has the same potential energy at all points along the equipotential surface.
  • We have to prove that at every point on an equipotential surface, the surface must be perpendicular to the electric field.

The potential between two points (A) and (B) on an equipotential surface is given by:

W AB = q ΔV = -q \int\limits^A_B E dS

By the definition  ΔV at an equipotential surface is zero.

-q \int\limits^A_B E dS = 0

\int\limits^A_B E dS = 0

\int\limits^A_B ES cos θ = 0

Therefore, cos θ <em> must be 0°</em> or 90° for a field to be non electric.

Hence, it is proved that at every point on an equipotential surface, the surface must be perpendicular to the electric field.

To learn more about equipotential surfaces, refer to:

brainly.com/question/14675095

#SPJ4

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A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
evablogger [386]

Answer:

a)n= 3.125 x 10^{19 electrons.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x 10^{-3 m

radius 'r' = d/2 => 1.025 x 10^{-3 m

no. of electrons 'n'= 8.5 x 10^{28}

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrons.

b)  the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) The typical speed'V_{d' of an electron is given by:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) According to these equations,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

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2 years ago
A cannon is fired horizontally at 243 m/s off of a 62 meter tall, shear vertical cliff. How far in meters from the base of the c
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Answer:

865.08 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 243 m/s

Height (h) of the cliff = 62 m

Horizontal distance (s) =?

Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:

Height (h) of the cliff = 62 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

62 = ½ × 9.8 × t²

62 = 4.9 × t²

Divide both side by 4.9

t² = 62/4.9

Take the square root of both side.

t = √(62/4.9)

t = 3.56 s

Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:

Initial velocity (u) = 243 m/s

Time (t) = 3.56 s

Horizontal distance (s) =?

s = ut

s = 243 × 3.56 s

s = 865.08 m

Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.

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