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3 years ago

Is it possible for the electric field between two positive charges to equal zero along the line joining the two charges?

Physics

1 answer:

MrRa [10]3 years ago

6 0

Answer:

Yes electric field between the line joining may be zero when both the charges are of opposite magnitude

Explanation:

Electric field due to a point charge is given by $E=\frac{1}{4\pi \epsilon _0}\frac{Q}{r^2}$

From the relation of magnitude of electric field we can see that magnitude of electric field depends upon the m,magnitude of charge and distance between them

So electric field between the line joining may be zero when both charges are of opposite magnitude ( Means if one charge is positive then other is negative )

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A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows:

$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

$0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}$

$v_{1,f} = 5.86-2.25\cdot v_{2,f}$

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

$13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}$

$13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}$

$13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}$

$2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0$

$2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0$

There are two solutions:

$v_{2,f} = 0\,\frac{m}{s}$ or $v_{2,f} = 3.606\,\frac{m}{s}$

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ( $v_{2,f} = 3.606\,\frac{m}{s}$ )

$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0

3 years ago

The partial negative charge in a molecule of water occurs because

just olya [345]

the electrons shared between the oxygen and hydrogen atoms spend more time around the oxygen atom nucleus than around the hydrogen atom nucleus

4 0

4 years ago

Identify two fields where physical quantities are used in motion calculations

larisa86 [58]

The two fields were physical quantities are used in motion calculations are length and mass with time.

The physical quantity in a field is referred as every point in a particular space time.

<h3>How physical quantities are used in motion calculations?</h3>

If we consider an object, the physical property of the object is considered as physical quantity and to measure that object is known as units. The Physical quantity can be classified as elemental physical quantity and derived physical quantity. Length, mass, time, etc.. are elemental physical quantity, momentum, density, acceleration, etc... are derived physical quantity. Only for charge and temperature the physical quantity will be less than zero.

Length, mass and time are the physical quantities used in motion calculations.

Learn more about motion calculations,

brainly.com/question/8701763

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4 0

2 years ago

What is <br>M=35g, V=17cm³

cestrela7 [59]

Answer: 25

Explanation:

6 0

3 years ago

The potential difference between the surface of a 3.0-cm-diameter power line and a point 1.0 m distant is 3.9 kV. Find the line

Korolek [52]

Answer:

The linear charge density is 5.19 X 10⁻⁶ C/m

Explanation:

The potential difference between two cylinders, is given as

V = (λ/2πε)ln(b/a)

where;

λ is the line charge density on the power line.

b is the distance between the power line = 1 m

a is the radius of the wire = 1.5 cm = 0.015 m

ε is the permittivity of free space = 8.9 X 10⁻¹² C

V*2πε = λ* ln(b/a)

3900 *(2π*8.9 x10⁻¹²)= λ *ln(1/0.015)

2.1812 X 10⁻⁷ = 4.1997* λ

λ = 5.19 X 10⁻⁶ C/m

Therefore, the linear charge density is 5.19 X 10⁻⁶ C/m

6 0

3 years ago