Question

You have two charges, a 1.7μC charge placed at the origin and a -3.6μC charge placed 2.0m to the right.  Determine where, on the line between the two charges, the electric potential difference will be zero.

Answer 1

Let the distance from the 1.7μC charge to the point where the electric potential difference is zero be $x$. Then, the distance from the -3.6μC charge to that point be $2-x$. Then, we can write:

$$V_{total}=V_1+V_2=0,$$ $$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0,$$ $$\dfrac{q_1}{x}+\dfrac{q_2}{2-x}=0,$$ $$q_1(2-x)=-q_2x,$$ $$x=\dfrac{2q_1}{q_1-q_2}=\dfrac{2\ m\cdot1.7\ \mu C}{1.7\ \mu C-(-3.6\ \mu C)}=0.64\ m.$$

The electric potential difference will be zero at a distance of 0.64 meters from the 1.7μC charge and 1.36 meters from the -3.6μC charge.