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xz_007 [3.2K]
2 years ago
9

A parallel plate capacitor fully charged to voltage V is connected to the battery (the voltage on the plates remains fixed). If

the plates are pulled away from each other, how would it affect the charge on the capacitor
Physics
1 answer:
Papessa [141]2 years ago
7 0

Charge will decreases.

A parallel plate capacitor when it is fully charged to voltage V is given as:

                    C = Q/V

The capacitance of parallel plate capacitor with two plates of Area A separated by distance d and no dielectric material between plates is

                    C = ε₀ A /d

since from above equation it shows C is proportional to Q and also C is inversely proportional to distance d.

So, ATQ when d increases C will decrease which in result decreases charge on the capacitor.

Thus,  Charge will decrease.

Learn more about capacitance here:

     brainly.com/question/17115454

          #SPJ4

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A positively charged particle initially at rest on the ground accelerates upward to 160 m/s in 2.10 s. The particle has a charge
mamaluj [8]

Answer:

The magnitude of the electric field is 8.6\times10^{2}\ N/C

Explanation:

Given that,

Time t = 2.10 s

Speed = 160 m/s

Specific charge =Ratio of charge to mass = 0.100 C/kg

We need to calculate the acceleration

Using equation of motion

a=\dfrac{v-u}{t}

Put the value into the formula

a=\dfrac{160-0}{2.10}

a=76.19\ m/s^2

We need to calculate the magnitude of the electric field

Using formula of electric field

E=\dfrac{F}{q}

E=\dfrac{ma}{q}

E=\dfrac{a+g}{\dfrac{q}{m}}

Put the value into the formula

E=\dfrac{76.19+9.8}{0.100}

E=8.6\times10^{2}\ N/C

The direction is upward.

Hence, The magnitude of the electric field is 8.6\times10^{2}\ N/C

4 0
3 years ago
Read 2 more answers
What is the Net Force?
marshall27 [118]

It is 800 N FN = 600N + 200 N = 800 N Answer to your question: The net force is all Newton's second law. It is the force that acts on a body or a particle. for example: It is the force we make when we push a car or something heavy that is in a straight line. .

3 0
3 years ago
Car A is traveling west at 40 mi/h and car B is traveling north at 40 mi/h. Both are headed for the intersection of the two road
Gwar [14]

Explanation:

It is given that,

    \frac{dx}{dt} = -40 mi/h,     \frac{dx}{dt} = -40 mi/h

The negative sign indicates that x and y are decreasing.

We have to find \frac{dz}{dt}. Equation for the given variables according to the Pythagoras theorem is as follows.

              z^{2} = x^{2} + y^{2}

Now, we will differentiate each side w.r.t 't' as follows.

        2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}

or,          \frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})

So, when x = 4 mi, and y = 3 mi then z = 5 mi.

As,       \frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})

                       = \frac{1}{5}(4 \times (-40) + 3 \times (-40))

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7 0
3 years ago
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sineoko [7]

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Explanation:

3 0
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I am trying to find the magnitude of a resultant vector. Do i take inconsideration the negatives when i find the x & y compo
attashe74 [19]
Absolutely !  If you have two vectors with equal magnitudes and opposite
directions, then one of them is the negative of the other.  Their correct
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(Think of fifty football players pulling on each end of the rope in a tug-of-war. 
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force on it is zero.)

This gross, messy explanation is completely applicable when you're totaling up
the x-components or the y-components.
4 0
3 years ago
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