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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 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m
ArbitrLikvidat [17]

Answer:

<h2>206.67N</h2>

Explanation:

The sum of force along both components x and y is expressed as;

\sum Fx = ma_x  \ and \ \sum Fy = ma_y

The magnitude of the net force which is also known as the resultant will be expressed as R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

a_x = \frac{d^2 x }{dt^2}

a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

7 0
3 years ago
As a projectile falls, what happens to the components of velocity?
netineya [11]

Answer:

Option (c).

Explanation:

An object when when projected at an angle, will have some horizontal velocity and vertical velocity such that,

v_x=v\cos\theta\ \text{and}\ v_y=v\sin\theta

\theta is the angle of projection

The horizontal component of the projectile remains the same because there is no horizontal motion. Vertical component changes at every point.

As a projectile falls, vertical velocity increases in magnitude, horizontal velocity stays the same .

7 0
3 years ago
Who water rocket starts from rest and roses straight up with an acceleration of 5 m/s until it runs out of water 2.5 seconds lat
Kitty [74]

Answer:

23. 4375 m

Explanation:

There are two parts of the rocket's motion

1 ) accelerating  (assume it goes upto  h1 height )

using motion equations upwards

s = ut+\frac{1}{2}*a*t^{2} \\h_1=0+\frac{1}{2}*5*2.5^{2} \\=15.625 m

Lets find the velocity after 2.5 seconds (V1)

V = U +at

V1 = 0 +5*2.5 = 12.5 m/s  

2) motion under gravity (assume it goes upto  h2 height )

now there no acceleration from the rocket. it is now subjected to the gravity

using motion equations upwards (assuming g= 10m/s² downwards)

V²= U² +2as

0 = 12.5²+2*(-10)*h2

h2 = 7.8125 m

maximum height = h1 + h2

                            = 15.625 + 7.8125

                            = 23. 4375 m

3 0
3 years ago
A karate master strikes a board with an initial velocity of 10.0 m/s, decreasing to 1.0 m/s as his hand passes through the board
kenny6666 [7]

The force exerted on the board by the karate master given the data is -4500 N

<h3>Data obtained from the question </h3>
  • Initial velocity (u) = 10 m/s
  • Final velocity (v) = 1 m/s
  • Time (t) = 0.002 s
  • Mass (m) = 1 Kg
  • Force (F) = ?
<h3>How to determine the force</h3>

The force exerted can be obtained as illustrated below:

F = m(v - u) / t

F = 1 (1 - 10) / 0.002

F = (1 × -9) / 0.002

F = -4500 N

Learn more about momentum:

brainly.com/question/250648

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2 years ago
Speed of light in air is 3 x 10 m/s, determine the speed of light in the glass fibre.
Dafna11 [192]
30 speed of light in the glass
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