Physics, 16.03.2020 19:28 masonbitterman7488

Two equal and opposite charges separated by a distance, d, experience an attractive force, F. If each charge is doubled in magnitude, what should be the new distance between them so that the force remains unchanged?

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If the each charge is doubled in magnitude and force remains unchanged then the distance between the charged particle is four times the original distance.

Explanation:

The force between two charged objects are shown by Coulomb's law.

In the scalar form, this law is represented as:

$F = k \frac{q_1 q_2}{d^2}$

where,

k = coulomb's constant

q₁ and q₂ are the magnitude of charged objects

d = distance between the charges.

According to the question,

Case1:

charge on object 1 = +q₁

Charge on object 2 = -q₂

Distance between the charges = d

$Force, F = k\frac{+q_1 X -q_2}{d^2} \\\\F = k\frac{-q_1q_2}{d^2}$

Case 2:

Charge on object 1 = +2q₁

Charge on object 2 = -2q₂

Force between the objects is same

New distance, dₓ = ?

If both the forces are equal then,

$k\frac{-q_1q_2}{d^2} = k\frac{+2q_1 X -2q_2}{d_x^2} \\\\$

On simplifying the equation we get,

$\frac{-1}{d^2} = \frac{-4}{d_x^2} \\\\d_x^2 = 4d^2$

So, if the each charge is doubled in magnitude and force remains unchanged then the distance between the charged particle is four times the original distance.

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

Answer from: Quest

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Two equal and opposite charges separated by a distance, d, experience an attractive force, F. If eac...