Electromagnetism - Capacitor

Two capacitors of capacitance 1.0 microfarad and 2.0 microfarad are each charged by being connected across a 5.0-volt battery. They are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together. What will be the magnitude of the final voltage across the 2.0-microfarad capacitor?
A. 0V
B. 0.6V
C. 1.7V
D. 3.3V
E. 5.0V

(GR9677 #62)

Solution:
Fig 1: Initial configuration.

The charges stored at each capacitor when they are connected across 5.0-V battery:



Fig 2: At the instant after reconnection (the plates of opposite charge are connected together)


Fig 3: In equilibrium

A short time later after reconnection, the net charge is conserved but redistributed between the two capacitors until the potential across them reaches equilibrium (the same).

Thus, the net charges on the upper and lower plates are:



And,


To find ,take the upper plate:




Answer: C

Electromagnetism - Electric Potential

A cube has a constant electric potential $V$ on its surface. If there are no charges inside the cube, the potential at the center of the cube is
A. zero
B. $V/8$
C. $V/6$
D. $V/2$
E. $V$

(GR8677 #52)

Solution:
Gauss’ Law: $\oint \bar{E} \cdot \hat{n} dA=\frac{q_{\textup{enc}}}{\varepsilon_0}$
With $q_{\textup{enc}}=0 \rightarrow E=0$

The potential is related to the electric field by $E=\nabla\varphi$

Since $E=0 \rightarrow \varphi$ must be constant, which is given in the problem by $V$

Since the potential function has to remain continuous its value everywhere, inside or at the surface the cube $\rightarrow \varphi$ at the center of the cube is also $V$

Answer: E

Electromagnetism - Gauss' Law

A sphere of radius $R$ carries charge density proportional to the square of the distance from the center: $\rho=Ar^2$, where $A$ is a positive constant. At a distance of $R/2$ from the center, the magnitude of the electric field is:
A.
B.
C. $AR^3/24\varepsilon_0$
D. $AR^3/5\varepsilon_0$
E. $AR^3/3\varepsilon_0$

(GR9677 #61)

Solution:

Gauss’ Law:
 
The net charge within the Gussian surface with $\rho$ as a function of $r$ (non-uniformly charged sphere):



$\rightarrow E=\frac{1}{\pi R^2 \varepsilon_0}\cdot \frac{A \pi R^5}{40}=\frac{AR^3}{40\varepsilon_0}$
Answer: B