Question

The two metal spheres in Figure Q26.9 are connected by a metal wire with a switch in the middle. Initially the switch is open. Sphere 1, with the larger radius, is given a positive charge. Sphere 2, with the smaller radius, is neutral. Then the switch is closed. Afterward, sphere 1 has charge $Q_1$, is at potential $V_1$, and the electric field strength at its surface is $E_1$. The values for sphere 2 are $Q_2,V_2,\mathrm{and}{E}_2$.

a. Is $V_1$larger than, smaller than, or equal to $V_2$? Explain.

b. Is $Q_1$larger than, smaller than, or equal to $Q_2$? Explain.

c. Is $E_1$ larger than, smaller than, or equal to $E_2$? Explain.

Short answer

a.${\mathit{V}}_{\mathbf{1}}\mathbf{=}{\mathit{V}}_{\mathbf{2}}$

b.${\mathit{Q}}_{\mathbf{1}}\mathbf{>}{\mathit{Q}}_{\mathbf{2}}$

c.${\mathit{E}}_{\mathbf{1}}\mathbf{<}{\mathit{E}}_{\mathbf{2}}$

Step-by-step solution

: Given information and formula used   

Given :

After the switch is closed,

Charge, potential and electric field strength of sphere 1 : $Q_1,V_1,E_1$

Charge, potential and electric field strength of sphere 1 : $Q_2,V_2,E_2$.

Theory used :

Two proportional laws are :

$V\propto \frac{Q}{r}$

$E\propto \frac{V}{r}$

Determining if V1 larger than, smaller than, or equal to V2 

(a) When the switch is closed, the charge is transferred from Sphere 1 to Sphere 2 until their potentials are equal, i.e.

${\mathit{V}}_{\mathbf{1}}\mathbf{=}{\mathit{V}}_{\mathbf{2}}$.

Determining if Q1 larger than, smaller than, or equal to Q2?

(b) Because $V\propto \frac{Q}{r}$, the smaller sphere must carry less charge to have the same potential, hence ${\mathit{Q}}_{\mathbf{1}}\mathbf{>}{\mathit{Q}}_{\mathbf{2}}$.

Determining if E1 larger than, smaller than, or equal to E2 

c. For the electric field on the sphere's surface, we know that $E\propto \frac{V}{r}$Because both spheres have the same potential, the sphere with the smaller radius $\frac{V}{r}$will have the greater ratio. i.e.

${\mathit{E}}_{\mathbf{1}}\mathbf{<}{\mathit{E}}_{\mathbf{2}}$.