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1 year ago

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

1 answer:

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.

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Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11

tiny-mole [99]

Answer:

The line charge density is $1.59\times10^{-4}\ C/m$

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

$V=EA$

$V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2$

$\lambda=\dfrac{V\times2\epsilon_{0}}{r}$

Where, r = radius

V = potential difference

Put the value into the formula

$\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}$

$\lambda=1.59\times10^{-4}\ C/m$

Hence, The line charge density is $1.59\times10^{-4}\ C/m$

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3 years ago

Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the

Oksi-84 [34.3K]

-- The net force on the box is 2N to the left.

-- The box will move to the left and accelerate to the left.

-- F=ma . a=F/m . a=(2N)/(4kg).

a = 0.5 m/s^2 to the left.

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2 years ago

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Rom4ik [11]

Hope this helps you.

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