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

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