Answer to Question #93023 in Electricity and Magnetism for norah

Question

three equal point charges Q are arranged at the corners of a square of side L. draw the situation. what is the potential at the fourth corner, taking V=0 at a great distance?

Answer 1

The electric potential generated by a point charge is given by:

$V=K\frac{Q}{r}$

Where:

• Coulomb constant $K$
• Electric charge $Q$
• Distance from the load to the vertex.$r$

All charges have the same magnitude but the distance varies.

• Distance of electric charge 1 to point P.

$r_{1}=L$

• The electric potential generated by charge 1 at point P is:

$V_{1}=K\frac{Q}{r_{1}}$

Now

$V_{1}=K\frac{Q}{L}$

• Distance of electric charge 2 to point P

The Pythagorean Theorem is applied.

$r_{2}=\sqrt{L^{2}+L^{2}}$

$r_{2}=\sqrt{2L^{2}}$

FInally $r_{2}=\sqrt{2}L$

Electric potential generated from charge 2 to point P

$V_{2}=K\frac{Q}{r_{2}}$

Now

$V_{2}=K\frac{Q}{\sqrt{2}L}$ rationalizing $V_{2}=K\frac{Q}{\sqrt{2}L}*\frac{\sqrt{2}}{\sqrt{2}}$

Finally $V_{2}=\sqrt{2}K\frac{Q}{2L}$

• Distance from charge 3 to point P

$r_{3}=L$

Electric potential generated by charge 3 at point P

$V_{3}=K\frac{Q}{r_{3}}$

Now

$V_{3}=K=\frac{Q}{L}$

The potential at point P is

$V_{P}=V_{1}+V_{2}+V_{3}$

Now

$V_{P}=K\frac{Q}{L}+\sqrt{2}K\frac{Q}{2L}+K\frac{Q}{L}$

Adding similar terms

$V_{P}=2K\frac{Q}{L}+\sqrt{2}K\frac{Q}{2L}$

Adding fractions

$V_{p}=\frac{4KQ+\sqrt{2}KQ}{2L}$

Factoring.

$V_{p}=\frac{4+\sqrt{2}}{2}\frac{KQ}{L}$