Determine the magnitude of the electric field at the surface of a lead-196 nucleus, which contains 82 protons and 114 neutrons. Assume the lead nucleus has a volume 196 times that of one proton and consider a proton to be a sphere of radius 1.20 ✕ 10⁻¹⁵ m.

2.4 × 10 ²¹ N/C

Explanation:

The lead-196 nucleus has 82 proton

Q ( charge on the nucleus ) = 82e where e = 1.602 × 10⁻¹⁹ C

Q = 82 × 1.602 × 10⁻¹⁹ C = 1.314 × 10⁻¹⁷ C

vp ( volume of proton ) = 4/3π r³

V, volume of lead nucleus = 4/3πR³

4/3πR³ = 196 × 4/3π r³

R = ∛(196r³) = 5.81 r = 5.81 ×  1.20 ✕ 10⁻¹⁵ m = 6.97 × 10⁻¹⁵ m

magnitude of the electric field = KQ/R² = 8.99 × 10⁹ ×  1.314 × 10⁻¹⁷ C  / (6.97 × 10⁻¹⁵ m )² = 2.4 × 10 ²¹ N/C

higher potential

explanation:

because it is in a high speed so it must move into a of high potential