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il63 [147K]
3 years ago
8

How does the electric field intensity vary with the increase of distance of the point from the centre of a charged conducting sp

here? Explain with graph.​
Physics
1 answer:
Nataly_w [17]3 years ago
4 0

Answer:

(i)  Electric field outside the shell:

For point r>R; draw a spherical gaussian surface of radius r.

Using gauss law, ∮E.ds=q0qend

Since E is perpendicular to gaussian surface, angle betwee E is 0.

Also E being constant, can be taken out of integral.

So, E(4πr2)=q0q

So, E=4πε01r2q

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Lakes and rivers can be formed landforms from water
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Stars fuse most of the hydrogen in the entire star before they die.<br> 2 poir<br> True<br> False
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Once all hydrogen is depleted the star begins to die
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A loop of wire lays flat on a horizontal surface in an area with uniform magnetic field directed vertically down. Magnetic field
zimovet [89]

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B. counterclockwise

Explanation:

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3 years ago
"what is the horse's average velocity during the first ten seconds of its gallop
vladimir1956 [14]

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, from the completed question, determine the distance (in meters) the horse covered in the first ten seconds of it's gallop and apply the formula below.

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6 0
3 years ago
A 2 kg rock is at the edge of a cliff 20 meters above a lake The rock becomes loose and falls toward the water below. Calculate
natima [27]

Answer:

The potential energy (P.E) at the top is 392 J

The kinetic energy (K.E) at the top is 0 J

The potential energy (P.E) at the halfway point is 196 J.

The kinetic energy (K.E) at the halfway point is 196 J.

Explanation:

Given;

mass of the rock, m = 2 kg

height of the cliff, h = 20 m

speed of the rock at the halfway point, v = 14 m/s

The potential energy (P.E) and kinetic energy (K.E) when its at the top;

P.E = mgh

P.E = (2)(9.8)(20)

P.E= 392 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the top of the cliff = 0

K.E = ¹/₂(2)(0)²

K.E = 0

The potential energy (P.E) and kinetic energy (K.E) at the halfway point;

P.E = mg(¹/₂h)

P.E = (2)(9.8)(¹/₂ x 20)

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where;

v is velocity of the rock at the halfway point = 14 m/s

K.E = ¹/₂(2)(14)²

K.E = 196 J.

4 0
2 years ago
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