Question

Multiple Choice Questions.

1) The number of protons in the nucleus of an electrically neutral atom is equal to _________.
A) the number of electrons surrounding the nucleus.
B) the number of neutrons in the nucleus.
C) the number of neutrons surrounding the nucleus.
D) the number of electrons in the nucleus.
E) None of the other choices is correct.
2) The main difference between conductors and insulators is in terms of _________.
A) valence or conduction electrons.
B) protons.
C) atoms.
D) neutrons.
E) None of the other choices is correct.
3) When the distance between the two charges is doubled, the force between them is ___________.
A) quadrupled.
B) reduced by a factor of 4.
C) doubled.
D) reduced by a factor of 2.
E) reduced by a factor of 3.
4) When the magnitude of both interacting charges is increased by a factor of 2, the electrical forces between these charges is _________
A) reduced by a factor of 3.
B) reduced by a factor of 2.
C) reduced by a factor of 4.
D) doubled.
E) quadrupled.
5) Two charges, Q1 and Q2, are separated by a certain distance R. If the magnitudes of the charges are halved, and their separation is also halved, then what happens to the electrical force between these charges?
A) It is doubled.
B) It remains the same.
C) It increases by a factor of 16.
D) It is quadrupled.
E) It increases by a factor of 8.

Answer 1

(1) A) the number of electrons surrounding the nucleus.

(2) A) valence or conduction electrons.

(3) B) reduced by a factor of 4.

(4) E) quadrupled

(5) B) It remains the same.

Step-by-step explanation:

(1) The number of protons in the nucleus of an electrically neutral atom is equal to _________

protons are found in the nucleus while electrons surround the nucleus

A) the number of electrons surrounding the nucleus.

2) The main difference between conductors and insulators is in terms of _________.

conductors have mobile or free electrons while insulators have none.

A) valence or conduction electrons.

3) When the distance between the two charges is doubled, the force between them is ___________.

According to Coulomb's law;

`F = (Kq_1q_2)/(r^2)`

where;

F is the force between the charges

q₁ and q₂ are the two charges

r is the distance between the two charges

k is coulomb's constant

`F = (C)/(r^2) , \ C = kq_1q_2\\\\F_1r_1^2 = F_2r_2^2, \ But \ r_2 = 2r_1\\\\F_2=(F_1r_1^2)/(r_2^2) = (F_1r_1^2)/((2r_1)^2)= (F_1r_1^2)/(4r_1^2)\\\\F_2 = (F_1)/(4)`

B) reduced by a factor of 4.

4) When the magnitude of both interacting charges is increased by a factor of 2, the electrical forces between these charges is _________

Apply Coulomb's law again;

`F =(kq_1q_2)/(r^2) \\\\F = Cq_1q_2, \ C = (k)/(r^2) \\\\(F_1)/((q_1q_2)_1) = (F_2)/((q_1q_2)_2) \\\\But \ (q_1q_2)_2 = (2q_12q_2)_1 = 4(q_1q_2)_1\\\\(F_1)/((q_1q_2)_1) = (F_2)/(4(q_1q_2)_1) \\\\F_2 = 4 F_1`

E) quadrupled

5) Two charges, Q1 and Q2, are separated by a certain distance R. If the magnitudes of the charges are halved, and their separation is also halved, then what happens to the electrical force between these charges?

Apply Coulomb's law;

`F = (kQ_1Q_2)/(R^2) \\\\k = (FR^2)/(Q_1Q_2) \\\\(F_1R^2)/(Q_1Q_2) = (F_2(R/2)^2)/((Q_1/2)(Q_2/2)) \\\\(F_1R^2)/(Q_1Q_2) =(F_2((1)/(4)R^2 ))/((1)/(4) (Q_1Q_2))\\\\F_2 = F_1`

B) It remains the same.