Question

A neutron consists of one 'up' quark of charge +2e/3 and two "down" quarks each having charge of -e/3.if we assume that the down quarks of 2.6*10^-15m a part inside the neutron what is the magnitude of the electrostatic force between them ? Expained it​

Answer 1

ANSWER:

• The magnitude of the electrostatic force between the two down quarks = 3.79 N.

GIVEN:

• A neutron consists of one up quark of charge + 2e/3 and two down quarks each having charge of - e/3.

• The down quarks are 2.6 × 10⁻¹⁵ m apart inside the neutron.

TO FIND:

• The magnitude of the electrostatic force between the two down quarks.

EXPLANATION:

$\red \clubs \ \boxed{\bold{\large{\pink{F=\dfrac{1}{4\pi\epsilon_o}\dfrac{q^2}{r^2}}}}}$

$\sf \dashrightarrow \dfrac{1}{4\pi\epsilon_o} = 9 \times {10}^{9} \ Nm^2C^{-2}$

$\sf \dashrightarrow q = - \dfrac{e}{3}\ C$

$\sf \dashrightarrow r = 2.6 \times 10^{-15}\ m$

$\sf \dashrightarrow F=9 \times {10}^{9} \times \dfrac{ \bigg(\dfrac{e}{3} \bigg)^2}{(2.6 \times 10^{-15})^2}$

$\sf \dashrightarrow F=\dfrac{ 9 \times {10}^{9} \times e^2}{9(2.6 \times 10^{-15})^2}$

$\sf \dashrightarrow F=\dfrac{{10}^{9} \times e^2}{6.76 \times 10^{-30}}$

$\sf \dashrightarrow F=\dfrac{{10}^{39} \times e^2}{6.76}$

$\bf We\ know\ that\ e = 1.6 \times 10^{-19}\ C$

$\sf \dashrightarrow F=\dfrac{{10}^{39} \times (1.6 \times 10^{-19})^2}{6.76}$

$\sf \dashrightarrow F=\dfrac{{10}^{39} \times 2.56 \times 10^{-38}}{6.76}$

$\sf \dashrightarrow F=\dfrac{2.56 \times 10}{6.76}$

$\sf \dashrightarrow F=\dfrac{25.6}{6.76}$

$\sf \dashrightarrow F=3.79\ N$

Hence the magnitude of the electrostatic force between the two down quarks = 3.79 N.

Answer 2

$\huge\color{black}{\underline{\underline{answer\::}}}$

F = 3.79N

$\huge\color{yellow}{\underline{\underline{explanation\::}}}$

Refers to the attachment....