458. A set of equipotential surfaces in a region has potential differences between adjacent surfaces as follows: 5 V, 10 V, and 15 V. The spacing between the first pair of surfaces is 2 cm. If the electric field strength is inversely proportional to the spacing between surfaces, what is the spacing between the third pair of surfaces?
Key Concept: Electric field strength, Equipotential surfaces
c) 6 cm
[Solution Description]
Given that $E = -\frac{dV}{dr}$, for the same potential difference, the spacing $dr$ is inversely proportional to $E$.
From the first pair: $E_1 = \left| \frac{dV_1}{dr_1} \right| = \frac{5 \, \text{V}}{0.02 \, \text{m}} = 250 \, \text{V/m}$.
For the second pair, since the potential difference doubles, but $E$ remains inversely proportional to $dr$, if $E_2$ changes, the new spacing will adjust accordingly.
However, for this problem, we assume $E$ is constant across regions. Hence, for the third pair with $\Delta V = 15 \, \text{V}$, using $dr_3 = \frac{\Delta V_3}{E_1} = \frac{15}{250} = 0.06 \, \text{m} = 6 \, \text{cm}$.
But this contradicts the initial condition of varying $E$. Instead, recognize that if spacing increases for higher $\Delta V$, then $dr_3 = dr_1 \times \frac{\Delta V_3}{\Delta V_1} = 2 \, \text{cm} \times \frac{15}{5} = 6 \, \text{cm}$.
Your Answer is correct.
c) 6 cm
[Solution Description]
Given that $E = -\frac{dV}{dr}$, for the same potential difference, the spacing $dr$ is inversely proportional to $E$.
From the first pair: $E_1 = \left| \frac{dV_1}{dr_1} \right| = \frac{5 \, \text{V}}{0.02 \, \text{m}} = 250 \, \text{V/m}$.
For the second pair, since the potential difference doubles, but $E$ remains inversely proportional to $dr$, if $E_2$ changes, the new spacing will adjust accordingly.
However, for this problem, we assume $E$ is constant across regions. Hence, for the third pair with $\Delta V = 15 \, \text{V}$, using $dr_3 = \frac{\Delta V_3}{E_1} = \frac{15}{250} = 0.06 \, \text{m} = 6 \, \text{cm}$.
But this contradicts the initial condition of varying $E$. Instead, recognize that if spacing increases for higher $\Delta V$, then $dr_3 = dr_1 \times \frac{\Delta V_3}{\Delta V_1} = 2 \, \text{cm} \times \frac{15}{5} = 6 \, \text{cm}$.