Answer:
The same number of atoms of each element must appear on both sides of a chemical equation. However, simply writing down the chemical formulas of reactants and products does not always result in equal numbers of atoms. You have to balance the equation to make the number of atoms equal on each side of an equation.
Explanation:
I hope thats what u needed.
Mass of an electron = 9.110 x 10⁻³¹ kg.
Mass of a proton = 1.6727 x 10⁻²⁷ kg
∴ mass of a proton/mass of an electron = 1.6727 x 10⁻²⁷ kg/9.110 x 10⁻³¹ kg.
~1836
∴ mass of a proton = 1836 x mass of an electron.
∴ mass of an electron is insignificant to the mass of an atom.
∴mass of an atom = mass of protons + mass of neutrons
Explanation:
The magnitude of the electric field between the plates is given by
E = -ΔV/d
minus sign indicates Potential decreases in the direction of electric field
where
ΔV is the potential difference between the plates
D is the distance between the plates.
The work done when carrying an electrical charge on an equipotential surface between one position to the other is zero W= q(V-V)=0 The electric field lines of force are always perpendicular to an equipotential surface. That conductor in an equipotential surface as direction E is at right angles to an eauipotential surface The intensity of the electric field along an equipotential surface is always zero. Equipotential surfaces never collide with each other as this would mean that at that point, there are two alternative values that are not true.
The initial force between the two charges is given by:

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:
1. F
In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.
So, we have:

So, the new force is:

So the force has not changed.
2. F/4
In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.
So, we have:

So, the new force is:

So the force has decreased by a factor 4.
3. 6F
In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.
So, we have:

So, the new force is:

So the force has increased by a factor 6.