Question

A scientist was examining the electrostatic interaction between charged spheres. The setup compromised two concentric spheres, inner one solid and outer one hollow, each charged with equal but opposite charges. Initially, he measured the potential difference between the two spheres and recorded it as $\phi$. Curious about the effect of altering the charges, he increased the charge on the hollow sphere to three times its original value. Now, he wonders how this change impacts the potential difference between the two spheres. What will the new potential difference be?

• A. -3$\phi$
• B. $\phi$
• C. 2$\phi$
• D. 9$\phi$

Answer 1

Answer: (B)

$\phi$

Answer 2

Let the inner sphere have radius $r_1$ and charge $q_1$, and the outer hollow sphere have radius $r_2$ and charge $q_2$. The potential difference between the inner and outer spheres is given by $\Delta V = V_{inner} - V_{outer}$. The potential at the surface of the inner sphere is $V_{inner} = \frac{1}{4\pi\epsilon_0} (\frac{q_1}{r_1} + \frac{q_2}{r_2})$. The potential at the surface of the outer sphere is $V_{outer} = \frac{1}{4\pi\epsilon_0} (\frac{q_1}{r_2} + \frac{q_2}{r_2})$. The potential difference is $\phi = V_{inner} - V_{outer} = \frac{1}{4\pi\epsilon_0} (\frac{q_1}{r_1} - \frac{q_1}{r_2})$.

Initially, the spheres have equal but opposite charges. Let $q_1 = q$ and $q_2 = -q$. So, the initial potential difference is $\phi = \frac{1}{4\pi\epsilon_0} (\frac{q}{r_1} - \frac{q}{r_2}) = \frac{q}{4\pi\epsilon_0} (\frac{1}{r_1} - \frac{1}{r_2})$.

The charge on the hollow sphere is increased to three times its original value. Original charge on the hollow sphere ($q_2$) was $-q$. The new charge on the hollow sphere ($q_2'$) is $3 \times (-q) = -3q$. The charge on the inner sphere ($q_1'$) remains $q$.

The new potential difference $\phi'$ is: $V'_{inner} = \frac{1}{4\pi\epsilon_0} (\frac{q}{r_1} + \frac{-3q}{r_2})$ $V'_{outer} = \frac{1}{4\pi\epsilon_0} (\frac{q}{r_2} + \frac{-3q}{r_2}) = \frac{-2q}{4\pi\epsilon_0 r_2}$ $\phi' = V'_{inner} - V'_{outer} = \left(\frac{1}{4\pi\epsilon_0} \frac{q}{r_1} - \frac{1}{4\pi\epsilon_0} \frac{3q}{r_2}\right) - \left(\frac{1}{4\pi\epsilon_0} \frac{q}{r_2} - \frac{1}{4\pi\epsilon_0} \frac{3q}{r_2}\right)$ $\phi' = \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{3}{r_2}\right) - \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r_2} - \frac{3}{r_2}\right)$ $\phi' = \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{3}{r_2} - \frac{1}{r_2} + \frac{3}{r_2}\right)$ $\phi' = \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2}\right)$

Thus, the new potential difference $\phi'$ is equal to the original potential difference $\phi$.