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Genrish500 [490]
3 years ago
7

An object that’s charged has more electrons than protons. An object that’s charged has fewer electrons than protons. An object t

hat’s charged has the same number of electrons and protons.
Physics
1 answer:
Levart [38]3 years ago
3 0
An element is neutral when the charged is equal to zero. This means that the number of electrons and protons are equal. When an object has a NEGATIVE charge, its electrons are more than its protons. Also, the object has a POSITIVE charge when the number of its electrons are fewer compared to the number of the protons. 
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Answer:

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2 years ago
How much work does the electric field do in moving a proton from a point with a potential of +145 v to a point where it is -55 v
Aleks04 [339]
The work done by the electric field in moving a charge is the negative of the potential energy difference between the two locations, which is the product between the magnitude of the charge q and the potential difference \Delta V:
W=-\Delta U=-q\Delta V=-q (V_f -V_i)
The proton charge is q=1.6 \cdot 10^{-19}C, and the two locations have potential of V_i = +145 V and V_f=-55 V, therefore the work is
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2 years ago
There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12
Licemer1 [7]

The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

        C = \frac{Q}{\Delta V}

        C = ε₀ \frac{A}{d}

we solve for the charge (Q)

        \frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

         Q = \epsilon_o \  \frac{A \ \Delta V_1 }{d_1}

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

         Q = \epsilon_o \  \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

          \frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}

          ΔV₂ = \frac{d_2}{d_1 } \ \Delta V_1

The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

          Em₀ = U = q DV₂

          Em₀ = q  \frac{d_2}{d_1 } \ \Delta V_1

           

final point. Proof load on the right plate

         Em_f = K

energy is conserved

         Em₀ = em_f

         q  \frac{d_2}{d_1 } \ \Delta V_1 = K

   

we calculate

         K = 1 10⁻⁹  12  \frac{0.005}{0.003}  

         K = 20 10⁻⁹ J

In this exercise, as the conditions at two different points of separation give, the area of ​​the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

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What type of friction prevents a pile of rocks from falling apart?
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A because it some type of friction
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