Tensors: Maxwell stress on an isotropic elastic material 
The Maxwell stress is $\sigma_{ij} = -\varepsilon_r[E_iE_j - 0.5E_kE_k\delta_{ij}]$. Assume that in a dielectric solid under study, the electric field vector   is aligned with the 1 axis. That is, $E_1 = E$ and $E_2 = E_3 = 0$. What is the Maxwell stress tensor for this solid and electric field given?  


It seems that the $0.5E_kE_k\delta_{ij}$ is never picked up due the kronecker delta. Is that correct?
$$                                                                        
      [\sigma_{ij}] =                                                           
      \begin{bmatrix}                                                           
        -\varepsilon E_1^2  & -\varepsilon E_1E_2 & -\varepsilon E_1E_3\\       
        -\varepsilon E_2E_1 & -\varepsilon E_2^2 & -\varepsilon E_2E_3\\        
        -\varepsilon E_3E_1 & -\varepsilon E_3E_2 & -\varepsilon E_3^2          
    \end{bmatrix}=
\begin{bmatrix}                                                           
        -\varepsilon E^2  & 0 & 0\\       
        0 & 0 & 0\\        
        0 & 0 & 0          
    \end{bmatrix} $$


Assume that the solid is made of an isotropic elastic material that deforms under the action of the Maxwell stress. Draw the material in the deformed state caused by the Maxwell stress. 


Since the solid is isotropic, wouldn't the block have the same orientation and dimension no matter the orientation?
 A: Notice the repeated index $k$, which implies summation. Thus, the expression $-0.5 E_k E_k \delta_{ij}$ is $-0.5 (E_1^2 + E_2^2 + E_3^2) \delta_{ij}$ = $-0.5 E^2\delta_{ij}$. 
This contributes $-0.5 E^2$ to the diagonal elements (where i=j and $\delta_{ij}=1$), and 0 to the off-diagonal elements. $E_i E_j$ is non-zero only in the (1,1) position of the matrix since you are given that $E_2=E_3=0$
So you have
$$
[\sigma_{ij}]=
-\varepsilon
\begin{bmatrix}
E^2 -0.5 E^2 & 0 & 0 \\
   0 & -0.5 E^2 & 0 \\
   0 &   0  &  -0.5 E^2 
\end{bmatrix}
=
\begin{bmatrix}
- 0.5 \varepsilon E^2 & 0 & 0 \\
   0 & 0.5 \varepsilon E^2 & 0 \\
   0 &   0  &  0.5 \varepsilon E^2 
\end{bmatrix}
$$
The second part of your question is more appropriate for PHYS.SE, but if you remember the stress-strain relation for an isotropic elastic material you should be able to sketch the deformed shape easily. Also, "isotropic" means (crudely) having the same material property in all directions, but that does not mean that the solid cannot deform at all! This is because strain is not a material property. 
More specifically, in an isotropic elastic material you can reduce the general stress-strain tensor relation $\sigma_{ij}=C_{ijkl}\epsilon_{kl}$ to one that involves only two constants. 
Edited to add:
Once you have the stress, you can use the strain-stress relation for an isotropic material to get the components of the strain. 
$$
\epsilon_{ij}=-\frac{\lambda \sigma_{kk} \delta_{ij} }{2\mu (3\lambda+2\mu)} + \frac{\sigma_{ij}}{2\mu}
$$
You will only get normal strain components if you work though this.
So a unit cube in the un-deformed configuration will change into a cuboidal solid of dimensions $(1+\epsilon_{11})(1+\epsilon_{22})(1+\epsilon_{33})$
