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JulsSmile [24]
2 years ago
15

Why must the test charge used to determine the electric field be very, very small?.

Physics
1 answer:
vekshin12 years ago
8 0

The test charge used to determine the electric field is very, very small so that its presence does not affect the electrical field because of the supply price. The electrical rate that produces the electric area is known as a source charge.

An electric-powered area is a physical area that surrounds electrically charged debris and exerts pressure on all differently charged particles inside the area, both attracting or repelling them. It also refers back to the physical field for a machine of charged particles.

Electric-powered discipline is described as the electrical force in keeping with unit fee. The path of the sector is taken to be the route of the force it might exert on a fine check price. the electric discipline is radially outward from a nice rate and radially in toward a bad point rate.

An electric discipline is a vicinity of space around an electrically charged particle or object in which an electric charge might sense force. An electric field is a vector amount and may be visualized as arrows going in the direction of or away from fees.

Learn more about  electric here brainly.com/question/24317284

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What is the device used to measure a potential difference in a circuit called?
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If the amplitude of a wave is increased the frequency of the wave will
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Remain the same

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A giant star has a luminosity of 300 its color is most likely to be
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3 years ago
A diffraction grating with 600 lines/mmlines/mm is illuminated with light of wavelength 510 nmnm. A very wide viewing screen is
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Answer:

A.2.95 m

B.7

Explanation:

We are given that

Diffraction grating=600 lines/mm

d=\frac{1 mm}{600}=\frac{1\times 10^{-3} m}{600}=1.67\times 10^{-6} m

Wavelength of light,\lambda=510 nm=510\times 10^{-9} m

l=4.6 m

A.We have to find the distance between the two m=1 bright fringes

sin\theta=\frac{m\lambda}{d}

For first bright fringe, =1

sin\theta=\frac{1\times 510\times 10^{-9}}{1.67\times 10^{-6}}=0.305

\theta=sin^{-1}(0.305)=17.76^{\circ}

The distance between two m=1 fringes

x=2ltan\theta=2\times 4.6 tan(17.76^{\circ})=2.95 m

Hence, the distance between two m=1 fringes=2.95 m

B.For maximum number of fringes,

sin\theta=1

sin\theta=\frac{m\lambda}{d}

Substitute the values

1=\frac{m\times 510\times 10^{-9}}{1.67\times 10^{-6}}

m=\frac{1.67\times 10^{-6}}{510\times 10^{-9}}=3.3\approx 3

Maximum number of bright fringes on the scree=2m+1=2(3)+1=7

8 0
3 years ago
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