Answer
(a) greater than
(b) less than
(c) $1.83\times 10^5N/C$
(d) $11.8^{\circ}$
Work Step by Step
(a) We know that $E_{net}\propto \frac{1}{r^2}$. The value of $E_{net}$ will be greater than that of the previous value. This is because all the charges are now close to point 3 than in the previous case.
(b) We know that the values of $\theta$ will be less than that of the previous value because $E_{net\space x}\lt E_{net\space y}$.
(c) We know that
$E_1=\frac{k|q_1|}{(\sqrt{(0.5m)^2+(0.25m)^2})^2}$
$\implies E_1=\frac{(8.99\times 10^9Nm^2/C^2)(2.9\mu C)}{(0.5m)^2(0.25m)^2}$
$\implies E_1=83.43\times 10^3N/C$
Similarly $E_2=\frac{(8.99\times 10^9Nm^2/C^2)(2.9\mu C)}{(0.5m)^2}$
$E_2=10.43\times 10^4N/C$
Now $E_{1x}=E_1cos45^{\circ}$
$\implies E_{1x}=(83.43\times 10^3N/C)(0.707)$
$E_{1x}=5.898\times 10^4N/C$
and $E_{1y}=E_1sin45^{\circ}$
$E_{1y}=(83.434\times 10^3N/C)(0.707)$
$E_{1y}=5.898\times 10^4N/C$
$E_{2x}=E_2cos 0^{\circ}$
$E_{2x}=(10.43\times 10^4N/C)(1)$
$E_{2x}=10.43\times 10^4N/C$
$E_{2y}=E_2sin0^{\circ}$
$E_{2y}=0$
Components $E_{net\space x}=E_{1x}+E_{2x}$
$E_{net\space x}=5.898\times 10^4N/C+10.43\times 10^4N/C$
$E_{net\space x}=5.898\times 10^4N/C+10.43\times 10^4N/C$
$E_{net\space x}=16.33\times 10^4N/C$
$E_{net\space y}=5.898\times 10^4N/C+0$
$E_{net\space y}=5.898\times 10^4N/C$
The magnitude of the net electric field can be determined as
$E{net}=\sqrt{E_{net\space}^2+E_{net\space y}^2}$
$\implies E_{net}=\sqrt{(16.33\times 10^4N/C)^2+(5.898\times 10^4N/C)^2}$
$E_{net}=1.83\times 10^5N/C$
(d) We know that
$\theta=tan^{-1}(\frac{E_{net\space y}}{E_{net\space y}})$
We plug in the known values to obtain:
$\theta=tan^{-1}(\frac{5.898\times 10^4N/C}{16.33\times 10^4N/C})$
$\theta=11.8^{\circ}$
