2. In the radioactive decay of 238 → 234Th + 4He, a 238 nucleus transforms to 234Th and an ejected 4He. (These are nuclei, not atoms, and thus electrons are not involved.) When the separation between 234Th and 4He is 9.0 × 10−15 m, what are the magnitudes of a) the electrostatic force between them and b) the acceleration of the 4He particle?

Explanation:

a )

charge on nucleus of 234Th = 1.6 x 10⁻¹⁹ x 90 ( atomic number of Th is 90 )

charge on 4 He = 1.6 x 10⁻¹⁹ x 2 ( in Helium , no of proton is 2 )

force of repulsion between them

= 9 x 10⁹ x 1.6 x 10⁻¹⁹ x 90 x 1.6 x 10⁻¹⁹ x 2  / ( 9 x 10⁻¹⁵ )²

= 51.2 x 10

= 512 N  .

b )

mass of 4 He

= 4 x 1.67 x 10⁻²⁷ kg

= 6.68 x 10⁻²⁷ kg

Acceleration

= Force / mass

= 512 / (6.68 x 10⁻²⁷)

= 76.64 x 10²⁷ m / s² .

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

i believe the answer is a