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Rufina [12.5K]
3 years ago
7

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

s. (a) if the two charges have the same sign, find the locationof all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points
Physics
1 answer:
NemiM [27]3 years ago
5 0

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is V=\frac{1}{4\pi\epsilon} \frac{q}{d}

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as-E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero  between the two charges,but we can find the points situated on the axial  line but  outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining  two charges  where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

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Trava [24]

Answer:

Their number should increase

Explanation:

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According to the accepted wave theory, light of any frequency will cause electrons to be emitted. Kinetic energy emitted by the electrons depends upon the intensity of light.

According to the accepted wave theory, number of electrons being ejected by the metal should increase

5 0
3 years ago
The process by which land remove surface materials is called
Citrus2011 [14]

Answer:

Deflation

Explanation:

The process by which wind removes surface materials is called deflation.

Deflation is a process by which wind erodes the Earth's exterior and the regions which experience severe erosion are called deflation zones.

deflation originates by the erosive force of a wind that removes the loosened area and this process is facilitated by a dry climate and a loss of vegetative cover that helps to lose the sediment.

deflation is common in the arid regions.

3 0
2 years ago
Read 2 more answers
Sound and light are both found as _____, with a variety of ____. The sun, a source of light waves specifically, releases a type
Novosadov [1.4K]

Answer:

Waves; wavelength; electromagnetic energy; ultraviolet light.

Explanation:

Sound are mechanical waves that are highly dependent on matter for their propagation and transmission.

Sound travels faster through solids than it does through either liquids or gases.

Light wave can be defined as an electromagnetic wave that do not require a medium of propagation for it to travel through a vacuum of space where no particles exist.

Hence, sound and light are both found as waves, with a variety of wavelength. The sun, a source of light waves specifically, releases a type of electromagnetic energy. It can be found as UVA or UVB types. These lights give off different levels of ultraviolet light, some of wich can be harmful.

Additionally, the ultraviolet spectrum is divided into three categories and these are; UVA, UVB and UVC.

6 0
2 years ago
The surface of the Earth changes from processe such as erosion. Which of these changes to Earth's surface is an example of erosi
Ad libitum [116K]

Answer:

D. the wind picking up dust and carrying it

Explanation:

Erosion is a process in which an agent transfer the top soil to another region, thereby exposing the lower soil. These agents have the ability to move the top layer of soil and deposit it at another place. The major agents in this case are; a running or flowing body of water and wind.

Therefore, the change to the Earth's surface that is an example of erosion is the wind picking up dust and carrying it. Thereby exposing the lower layers.

6 0
2 years ago
An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic
Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

R =u\cos\theta t

Put the value into the formula

\dfrac{230}{6} = u\cos\theta

u\cos\theta=38.33.....(I)

We need to calculate the height

Using vertical component

H=u\sin\theta t-\dfrac{1}{2}gt^2

Put the value in the equation

16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2

u\sin\theta=\dfrac{16+9.8\times18}{6}

u\sin\theta=32.06.....(II)

Dividing equation (II) and (I)

\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}

\tan\theta=0.8364

\theta=\tan^{-1}0.8364

\theta=39.90^{\circ}

(a). We need to calculate the initial speed

Using equation (I)

u\cos\theta\times t=38.33

Put the value into the formula

u =\dfrac{230}{6\times\cos39.90}

u=49.96\ m/s

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

5 0
3 years ago
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