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Nikolay [14]
3 years ago
8

Two charges +q and -q are situated at certain distance, at the point exactly midway betwee them what happens on electric field a

nd potential?
Physics
1 answer:
Luden [163]3 years ago
7 0

Answer: Electric field is not zero but potential is zero

Explanation:

In science, charge, also recognized as electric charge, electrical charge, or electrostatic charge and expressed q, is a component of a unit of body that reveals the extent to which it has more or fewer electrons than protons.

Considering the direction of the electric field is of positive to negative charge. Therefore, it determination be toward due −q also due to both the charges.

But for potential, it is a scalar quantity, so the sum of potential due to all the charges will be zero.

Henceforth, the option (c) Electric field is not zero but potential is zero is the correct answer.

a incorect

From the above statements, we could conclude that option (c) Electric field is not zero, but potential is zero is the correct answer. So, the option (a) Electric field and potential both are zero is an incorrect solution.

b incorect

Among all the options, the option (c) Electric field is not zero but potential is zero is the correct answer.

Therefore, the option (b) Electric field is zero but potential is not zero is an incorrect solution

d) is incorrect

It has been explained that the option (c) Electric field is not zero but potential is zero is the correct answer.

Hence, the option (d) Electric field is not zero but potential is zero is an incorrect solution.

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3 years ago
Which situation is contrary to Newton’s first law of motion?
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Answer:

An object at rest stays at rest as long as unbalanced forces act on it.

Explanation:

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4 0
3 years ago
Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W
jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

T^{2}=\frac{4\pi^{2}}{GM}r^{3}    (1)

Where;

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24}kg is the mass of the Earth

r  is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}

V_{X}=18224.783 m/s (9)

<u>For Satellite Y:</u>

V_{Y}=\sqrt{\frac{GM}{r_{Y}}} (10)

V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}

V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

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