The electric potential from point charges is $\displaystyle V=\frac{kq}{d}$.

Knowing that all three charges are identical, and knowing that the center point at which we are calculating the electric potential is equal distance from the charges, we can multiply the electric potential equation by three.

$\displaystyle V=3\frac{kq}{d}$

Plug in the given values and solve for $\displaystyle V$.

$\displaystyle V=3\frac{(8.99*10^9\ \frac{\text{Nm}^2}{\text{C}^2})(1* 10^{-6}\ \text{C})}{3*10^{-3}\ \text{m}}$

$\displaystyle V=8.99* 10^6\ \text{V}$

By the Pythagorean theorem, each charge is a distance

$\displaystyle d = \sqrt{\left (\frac{a}{2} \right )^2 + \left (\frac{a}{2} \right )^2+\left (\frac{a}{2} \right )^2} = \frac{\sqrt{3}}{2}a$

from the center of the cube, so the potential is

$\displaystyle V=8\times\left (\frac{Q}{4\pi\epsilon_0\left (\frac{\sqrt{3}}{2}a \right )} \right )$

$\displaystyle =\frac{4Q}{\pi\epsilon_0a\sqrt{3}}$

Use polar coordinates with the given surface charge density,$\displaystyle \sigma(r)=\frac{qa}{2\pi(r^2+a^2)^{3/2}}$ and area element $\displaystyle rd\theta d\phi$. Noting that a point $\displaystyle r$ from the origin is a distance $\displaystyle \sqrt{r^2+a^2}$ from the point of interest, we calculate the potential as follows, integrating with respect to $\displaystyle r$ from $\displaystyle 0$ to $\displaystyle \infty$.

$\displaystyle V=\frac{1}{4\pi\epsilon_0} \int_{0}^{2\pi} \int_{0}^{\infty}\left (\frac{qa}{2\pi(r^2+a^2)^{3/2}} \right ) \frac{rdrd\theta}{\sqrt{r^2+a^2}}$

$\displaystyle =\frac{1}{4\pi\epsilon_0} (2\pi) \left(\frac{qa}{2\pi} \right )\int_{0}^{\infty}\left (\frac{1}{(r^2+a^2)^{3/2}} \right ) \frac{rdr}{\sqrt{r^2+a^2}}$

$\displaystyle =\frac{qa}{4\pi\epsilon_0} \int_{0}^{\infty}\frac{r}{(r^2+a^2)^2} dr$

$\displaystyle =\frac{qa}{4\pi\epsilon_0} \left [\frac{1}{2a^2}\left (\frac{x^2}{x^2+a^2} \right ) \right ]_{0}^{\infty}$ 

$\displaystyle =\frac{q}{8\pi\epsilon_0a} (1-0)$  (by limit)

$\displaystyle =\frac{q}{8\pi\epsilon_0a}$

Remark: This is exactly the charge distribution that would be induced on an infinite sheet of (grounded) metal if a negative charge $\displaystyle -q$ were held a distance $\displaystyle a$ above it.

Use spherical coordinates with the given surface charge density $\displaystyle \sigma(\theta)=b\cos^3\theta$, and area element $\displaystyle R^2\sin\theta d\theta d\phi$. Every point on the hemispherical shell is a distance $\displaystyle R$ from the origin, so we calculate the potential as follows, noting the limits of integration for $\displaystyle \theta$ range from $\displaystyle 0$ to $\displaystyle \pi/2$.

$\displaystyle V=\frac{1}{4\pi\epsilon_0}\int_{0}^{2\pi}\int_{0}^{\pi/2}(b\cos^3\theta)\left (\frac{1}{R} \right )R^2\sin\theta d\theta d\phi$

$\displaystyle =\frac{Rb}{4\pi\epsilon_0} (2\pi) \int_{0}^{\pi/2}\cos^3\theta\sin\theta d\theta$

$\displaystyle =\frac{Rb}{2\epsilon_0} \left[\frac{-\cos^4\theta}{4} \right]_{0}^{\pi/2}$

$\displaystyle =\frac{Rb}{8\epsilon_0}$

Draw a line from the center perpendicular to any side of the triangle. This line divides the side into two equal pieces of length $\displaystyle \frac{a}{2}$. From the center, draw another line to one of the vertices at the end of this side. This produces a 30-60-90 triangle with longer leg $\displaystyle \frac{a}{2}$, so the hypotenuse (the distance from the vertex to the center) is $\displaystyle \frac{a}{\sqrt{3}}$. The potential at the center is due to three of these charges, so it must be

$\displaystyle V=3\times \left ( \frac{Q}{4\pi\epsilon_0\left(\frac{a}{\sqrt{3}} \right )} \right)$

$\displaystyle =\frac{3\sqrt{3}Q}{4\pi\epsilon_0a}$

Use a polar coordinate system with surface charge density $\displaystyle \frac{Q}{\pi(R_2^2-R_1^2)}$ and area element $\displaystyle Rdrd\theta$. The distance from the point of interest to a point a distance $\displaystyle r$ from the center is $\displaystyle \sqrt{r^2+a^2}$, so the potential is

$\displaystyle V=\frac{1}{4\pi\epsilon_0}\int_{0}^{2\pi}\int_{R_1}^{R_2}\left(\frac{Q}{\pi(R_2^2-R_1^2)}\right )\frac{rdrd\theta}{\sqrt{r^2+a^2}}$

$\displaystyle =\frac{1}{4\pi\epsilon_0}(2\pi)\left(\frac{Q}{\pi(R_2^2-R_1^2)}\right )\int_{R_1}^{R_2}\frac{r}{\sqrt{r^2+a^2}}dr$

$\displaystyle =\frac{1}{2\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left[\sqrt{r^2+a^2} \right ]_{R_1}^{R_2}$

$\displaystyle =\frac{1}{2\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left(\sqrt{R_2^2+a^2} -\sqrt{R_1^2+a^2}\right )$

By the Pythagorean theorem, the distance from point P to charge $\displaystyle 2Q$ is $\displaystyle 5a$. Because point P is also $\displaystyle 3a$ from charge $\displaystyle -Q$, it follows that the potential is

$\displaystyle V=\frac{1}{4\pi\epsilon_0}\left( \frac{2Q}{5a}-\frac{Q}{3a}\right )$

$\displaystyle =\frac{Q}{60\pi\epsilon_0a}$

Use a polar coordinate system, the given linear charge density $\displaystyle \lambda(\theta)=b\sin^2\theta$, and length element $\displaystyle Rd\theta$. Since every point on the ring is the same distance $\displaystyle R$ from the center, we calculate the potential as

$\displaystyle V = \frac{1}{4\pi\epsilon_0}\int_{0}^{2\pi}b\sin^2\theta\left (\frac{1}{R} \right )R d\theta$

$\displaystyle =\frac{b}{4\pi\epsilon_0}\int_{0}^{2\pi}\sin^2\theta d\theta$

$\displaystyle =\frac{b}{4\pi\epsilon_0}\int_{0}^{2\pi}\frac{1-\cos2\theta}{2} d\theta$

$\displaystyle =\frac{b}{4\pi\epsilon_0}\left [\frac{\theta}{2}-\frac{\sin2\theta}{4} \right ]_{0}^{2\pi}$

$\displaystyle =\frac{b}{4\epsilon_0}$

By the law of cosines, the distance from the point to charge $\displaystyle +q$ is

$\displaystyle \sqrt{R^2+\left(\frac{a}{2} \right )^2 - 2R\left(\frac{a}{2} \right )\cos\theta}$ .

The distance to charge $\displaystyle -q$ can be found by using the law of cosines using the supplementary angle $\displaystyle \pi-\theta$, for which $\displaystyle \cos(\pi-\theta) = \cos(\pi)\cos\theta - \sin(\pi)\sin\theta = -\cos\theta$. Therefore the distance to $\displaystyle -q$ is

 $\displaystyle \sqrt{R^2+\left(\frac{a}{2} \right )^2 - 2R\left(\frac{a}{2} \right )(-\cos\theta)}$.

Lastly, the exact potential is given by

$\displaystyle \frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 -Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 + Ra\cos\theta}} \right )$.

Remark: Far from the dipole (approximating $\displaystyle R\gg a$) gives the much simpler equation for the potential of an ideal electric dipole $\displaystyle \frac{qa\cos\theta}{4\pi\epsilon_0 R^2}$.

Use a spherical coordinate system and place the point of interest a distance $\displaystyle a$ from the center on the z-axis. By the law of cosines, the distance from this point to any point on the sphere is $\displaystyle \sqrt{R^2 + a^2 -2Ra\cos\theta}$. Using surface charge density $\displaystyle \frac{Q}{4\pi R^2}$ and area element $\displaystyle R^2\sin\theta d\theta d\phi$, we evaluate the potential as:

$\displaystyle V =\frac{1}{4\pi\epsilon_0} \int_{0}^{2\pi}\int_{0}^{\pi}\left (\frac{Q}{4\pi R^2} \right )\frac{R^2\sin\theta d\theta d\phi}{\sqrt{R^2 + a^2-2Ra\cos\theta}}$

.$\displaystyle =\frac{1}{4\pi\epsilon_0} \left (2\pi \right )\left (\frac{Q}{4\pi} \right )\int_{0}^{\pi}\frac{\sin\theta d\theta d\phi}{\sqrt{R^2 + a^2-2Ra\cos\theta}}$

$\displaystyle =\frac{1}{4\pi\epsilon_0} \left (\frac{Q}{2} \right ) \left [\frac{1}{Ra}\sqrt{R^2 + a^2-2Ra\cos\theta} \right ]_{0}^{\pi}$

$\displaystyle =\frac{1}{4\pi\epsilon_0} \left (\frac{Q}{2} \right ) \frac{1}{Ra}\left (|R+a|-|R-a| \right )$

$\displaystyle =\frac{1}{4\pi\epsilon_0} \left (\frac{Q}{2} \right ) \frac{1}{Ra}(2R)$

$\displaystyle =\frac{Q}{4\pi\epsilon_0a}$

 Remarkably, this is the same potential that would exist a distance $\displaystyle a$ from a point charge.