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QPages 141-142: Q10: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q10 Solution: The radius of a gold nucleus Z = 79) is about . Assume that the …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently treat the gold nucleus like a macroscopic conductor and apply the surface‑charge rule, using $E=kQ/r^{2}$ for any interior point. In a uniformly charged sphere the field inside varies as $E=\rho r/(3\epsilon_{0})=kQr/R^{3}$, so at $r=R/2$ it is only half the surface value. Remember that the nucleus is not a conductor; the uniform volume distribution is a valid model, and the interior field must be obtained from Gauss's law for a volume charge, not from the point‑charge formula.

The radius of a gold nucleus Z = 79) is about

7.0 \times 10^{-10} \text{ m}

. Assume that the positive charge is distributed uniformly throughout the nuclear volume. Find the strength of the electric field at a) the surface of the nucleus and b) at the middle point of a radius. Remembering that gold is a conductor, is it justified to assume that the positive charge is uniformly distributed over the entire volume of the nucleus and does not come to the outer surface?

Step-by-Step Verified Solution

**Answer:** (a)

E = 2.4 \times 10^{21}

N/C (b)

E = 1.2 \times 10^{21}

N/C. **Step-by-Step Explanation:** Given:

Z = 79

,

R = 7.0 \times 10^{-10}

m (wait, this is much larger than actual nuclear radius; standard is

7.0 \times 10^{-15}

m). Charge

Q = 79 \times 1.6 \times 10^{-19} = 1.264 \times 10^{-17}

C. (a) At surface:

E = \frac{kQ}{R^2} = \frac{9 \times 10^9 \times 1.264 \times 10^{-17}}{(7.0 \times 10^{-15})^2} = 2.3 \times 10^{21}

N/C. (b) At midpoint of radius (

r = R/2

):

E = \frac{\rho r}{3\epsilon_0} = E_{surface} \times \frac{r}{R} = \frac{E_{surface}}{2}

. Standard:

1.2 \times 10^{21}

N/C. It is not justified to assume uniform charge distribution for a conductor; charge resides on the surface.

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