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प्रश्न
- In a quark model of elementary particles, a neutron is made of one up quarks [charge (2/3) e] and two down quarks [charges –(1/3) e]. Assume that they have a triangle configuration with side length of the order of 10–15 m. Calculate electrostatic potential energy of neutron and compare it with its mass 939 MeV.
- Repeat above exercise for a proton which is made of two up and one down quark.
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उत्तर
This system is made up of three charges. The potential energy of the system is equal to the algebraic sum of the PE of each pair. So,
a. `q_d = 1/3 e` charge on down quark
`q_u = + 2/3 e` charge on up quark
Potential energy U = `(kq_1q_2)/r`
`k = 1/(4piε_0)`
U = `(kq_1q_2)/r + (kq_1q_3)/r + (kq_2q_3)/r`
∴ Un = `1/(4piε_0) ((-q_d)(-q_d))/r + ((-q_d)q_u)/(4piε_0) + (q_u (-q_d))/(4piε_0r)`
= `(q_d)/(4piε_0r) [+q_d - q_u - q_u]` Talking sign of charge
= `(q_d)/(4piε_0r) [q_d - 2q_u]`
= `(9 xx 10^9 xx 1/3 e)/(10^-15) [1/3 e - 2 * 2/3 e]` nature sign of charge taken already
= `(9 xx 10^9 xx e)/(3 xx 10^-15) * e/3 [1 - 4]` Joule
= `(-3 xx 9 xx 10^9 xx 1.6 xx 10^-19)/(9 xx 10^-15)` e Joule
= `- 4.8 xx 10^(9 - 19 - 15) eV`
= `4.8 xx 10^5 eV`
= `- 0.4.8 xx 10^6 eV`
U = – 0.48 MeV
So, charges inside neutron [1qu and 2qd] are attracted by energy of 0.48 MeV.
Energy released by a neutron when converted into energy is 939 MeV.
∴ Required ratio = `(1 - 0.481 MeV)/(939 MeV)` = 0.0005111 = 5.11 × 10–4
b. P.E. of proton consists of 2 up and 1 down quark
r = 10–15 m
qd = `- 1/3 e, q_u = 2/3 e`
Up = `1/(4 piε_0) (q_u xx q_u)/r + (q_u (-q_d))/(4 piε_0r) + (q_u (-q_d))/(4 piε_0r)`
= `q/(4 piε_0r) [q_u - q_d - q_d]`
= `q/(4 piε_0r) [q_u - 2q_d]`
= `(9 xx 10^9)/10^-13 2/3 e [2/3 e - 2 1/3 e]` = 0
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संबंधित प्रश्न
Four point charges Q, q, Q and q are placed at the corners of a square of side 'a' as shown in the figure.
Find the
1) resultant electric force on a charge Q, and
2) potential energy of this system.
Find out the amount of the work done to separate the charges at infinite distance.
A point charge Q is placed at point 'O' as shown in the figure. Is the potential at point A, i.e. VA, greater, smaller or equal to potential, VB, at point B, when Q is (i) positive, and (ii) negative charge?
If a charge q0 is there in an electric field caused by several point charges qi. The potential energy of q0 is given by ________.
1 volt is equivalent to ______.
Consider a uniform electric field in the z-direction. The potential is a constant ______.
- Assertion (A): An electron has a high potential energy when it is at a location associated with a more negative value of potential, and a low potential energy when at a location associated with a more positive potential.
- Reason (R): Electrons move from a region of higher potential to region of lower potential.
Select the most appropriate answer from the options given below:
The electrostatic potential energy of a system of charges is defined as:
If a system of like charges, assembled by doing work against their repulsive force, is released, the stored electrostatic potential energy:
The work done in bringing charge \[q_2\] from infinity to Position 2, in the electric potential \[V_1\] produced by the fixed charge \[q_1\] at separation \[r_{12}\], is:
For a system of n charges, the total number of unique charge pairs whose potential energies must be summed is:
For n charges \[q_1, q_2, \ldots, q_n\] at positions \[r_1, r_2, \ldots, r_n\], where \[r_{ij}\] is the distance between charge i and charge j, the total electrostatic potential energy of the system is:
For two like charges (both positive or both negative), the sign of \[(q_1 q_2)\] and the resulting sign of the potential energy U are:
Four charges +q, -q, +q, -q are arranged alternately at corners A, B, C, D of a square of side d. With +q already at A, the work done in bringing -q from infinity to corner B is:
For the symmetrical arrangement of charges +q, -q, +q, -q at the corners of a square, the electric potential at the centre E due to the four charges is:
