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3 years ago

How does the electrical force relate to the charge of an object?

Physics

2 answers:

shtirl [24]3 years ago

6 0

Answer:

It is directly proportional to the charge

Explanation:

The magnitude of the electrical force on a charge is given by:

$F=qE$

where

q is the magnitude of the charge

E is the magnitude of the electric field acting on the charge

From the formula, we see that F (the force) is directly proportional to q /the magnitude of the charge), so the correct answer is

It is directly proportional to the charge

We can also add something about the direction of the force:

- when the charge is positive, the force has positive sign, which means that its direction is the same as the direction of the electric field (repulsive force)

- when the charge is negative, the force has negative sign, which means that its direction is opposite to the direction of the electric field (attractive force)

mezya [45]3 years ago

4 0

The electric force between two object is directly proportional to the net charge on each object and inversely proportional to the square of the distance between them. So, the answer is: The electrical force is directly proportional to the charge.

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A camera is equipped with a lens with a focal length of 34 cm. When an object 2.4 m (240 cm) away is being photographed, what is

puteri [66]

Answer:

The magnification is -6.05.

Explanation:

Given that,

Focal length = 34 cm

Distance of the image =2.4 m = 240 cm

We need to calculate the distance of the object

$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$

Where, u = distance of the object

v = distance of the image

f = focal length

Put the value into the formula

$\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}$

$\dfrac{1}{u}=\dfrac{103}{4080}$

$u =\dfrac{4080}{103}$

The magnification is

$m = \dfrac{-v}{u}$

$m=\dfrac{-240\times103}{4080}$

$m = -6.05$

Hence, The magnification is -6.05.

6 0

2 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

3 years ago

A plant blossoms with violet-colored flowers. The flowers appear violet because they absorb all light rays except for____ rays.

jolli1 [7]

~Hello there! ^_^

Your question: A plant blossoms with violet-colored flowers. The flowers appear violet because they absorb all light rays except for____ rays.

Your answer: A plant blossoms with violet-colored flowers. The flowers appear violet because they absorb all light rays except for violet rays.

Hope this helps~

5 0

2 years ago

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elixir [45]

Answer:

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Explanation:

3 0

2 years ago

Is visible light a form of radiation? Explain.

natali 33 [55]

Answer:

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Explanation:

6 0

2 years ago