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larisa [96]
2 years ago
10

Fill in the blank. Consider the inverse square law: When light leaves a light bulb, it spreads out over more and more space as i

t goes. This makes the light thinner, with less and less light present the further from the light bulb we look. If we stand looking at a light bulb and see how bright it is, then move to be four times farther away from the light bulb, the light from the bulb will look ____________ less bright. Group of answer choices
Physics
1 answer:
aliya0001 [1]2 years ago
8 0

Answer:

Explanation:

Intensity of light is inversely proportional to distance from source

I ∝ 1 /r²  where I is intensity and r is distance from source . If I₁ and I₂ be intensity at distance r₁ and r₂ .

I₁ /I₂ = r₂² /r₁²

If r₂ = 4r₁ ( given )

I₁ / I₂ = (4r₁ )² / r₁²

= 16 r₁² / r₁²

I₁ / I₂ = 16

I₂ = I₁ / 16

So intensity will become 16 times less bright .

"16 times " is the answer .

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(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
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Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

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The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

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