An electron is to be accelerated from $3.00\times {10}^6$ m/s to $8.00\times {10}^6$ m/s. Through what potential difference must the electron pass to accomplish this?

Two point charges $q_1 = +2.40$ nC and $q_2 = -6.50$ nC are $0.100$ m apart. Point $A$ is midway between them; point $B$ is $0.080$ m from $q_1$ and $0.060$ m from $q_2$ (Fig. E$23.19$). Take the electric potential to be zero at infinity. Find the potential at point $B$.

Two point charges of equal magnitude $Q$ are held a distance $d$ apart. Consider only points on the line passing through both charges. If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?).

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are $4.98$ V and $16.2$ V/m, respectively. (Take $V = 0$ at infinity.) What is the distance to the point charge?

Two point charges $q_1=+2.40$ nC and $q_2=-6.50$ nC are $0.100$ m apart. Point $A$ is midway between them; point $B$ is $0.080$ m from $q_1$ and $0.060$ m from $q_2$ (Fig. E$23.19$). Take the electric potential to be zero at infinity. Find the potential at point $A$.

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are $4.98$ V and $16.2$ V/m, respectively. (Take $V=0$ at infinity.) What is the magnitude of the charge?

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are $4.98$ V and $16.2$ V/m, respectively. (Take $V = 0$ at infinity.) Is the electric field directed toward or away from the point charge?