Question

Two-point charges q₁ = +2.40 nC and q₂ = -6.50 nC are 0.100 m apart. Point A is midway between them; point B is 0.080 m from q₁ and 0.060 m from q₂. Take the electric potential to be zero at infinity. Find

(a) the potential at point A;

(b) the potential at point B;

(c) the work done by the electric field on a charge of 2.50 nC that travels from point B to point A.

Short answer

a) The potential at A due to both charges is -738.0 volts.

b) The potential at B due to both charges is -705.0 volts.

c) The work done by the electric field on a charge that travels from B to A equals 8.30 10^-8 Joules.

Step-by-step solution

Basic definition

An electric field is the physical field that surrounds a charge and exerts a force on other charged particles in the field. Mathematically, it is defined as the electric force per unit charge.

Electric potential is defined as the amount of work done in moving a unit charge from a reference point to a point against the electric field of another charge. Mathematically, it is equal to potential energy per unit charge.

If a charge is positive, then the potential created at a point by it is also positive, but if it is negative, then the potential created by it is also negative.

Potential at Point A

(a)

Electric potential is equal to the potential energy per unit charge,

$V=\frac{U}{q_0}=k\frac{q}{r}$

Where k is coulomb’s constant, and r is the distance from the point charge to where potential is required.

Potential at point A:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}={\mathrm{V}}_{\mathrm{q}1}+{\mathrm{V}}_{{\mathrm{q}}_2} \\ =\mathrm{k}\frac{{\mathrm{q}}_1}{{\mathrm{r}}_1}+\mathrm{k}\frac{{\mathrm{q}}_2}{{\mathrm{r}}_2} \\ =\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(\frac{2.4\times {10}^{-9}\mathrm{C}}{0.05\mathrm{m}}-\frac{6.5\times {10}^{-9}\mathrm{C}}{0.05\mathrm{m}}\right) \\ =-738.0\mathrm{V} \end{gathered}$$

Therefore, the potential at A due to both charges is -738.0 volts.

Potential at Point B

(b)

Electric potential is equal to the potential energy per unit charge,

$V=\frac{U}{q_0}=k\frac{q}{r}$

Where k is coulomb’s constant, and r is the distance from the point charge to where potential is required.

Potential at point B:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{B}}={\mathrm{V}}_{{\mathrm{q}}_1}+{\mathrm{V}}_{{\mathrm{q}}_2} \\ =\mathrm{k}\frac{{\mathrm{q}}_1}{{\mathrm{r}}_1}+\mathrm{k}\frac{{\mathrm{q}}_2}{{\mathrm{r}}_2} \\ =\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}\right)\left(\frac{2.4\times {10}^{-9}\mathrm{C}}{0.08}-\frac{6.5\times {10}^{-9}\mathrm{C}}{0.06\mathrm{m}}\right) \\ =-705.0\mathrm{V} \end{gathered}$$

Therefore, the potential at B due to both charges is -705.0 volts.

Determination of Work done

(c)

Electric potential is equal to the potential energy per unit charge,

$$\begin{gathered} V=\frac{U}{q_0} \\ U=V{q}_0 \end{gathered}$$

Work done by an electric field on a charge that travels from B to A is equal to a change in its potential energy:

$$\begin{gathered} \mathrm{W}=-∆\mathrm{U} \\ =-{\mathrm{U}}_{\mathrm{A}}+{\mathrm{U}}_{\mathrm{B}} \\ =-{\mathrm{V}}_{\mathrm{A}}{\mathrm{q}}_0+{\mathrm{V}}_{\mathrm{B}}{\mathrm{q}}_0 \\ ={\mathrm{q}}_0\left(-{\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}\right) \\ =\left(2.5\times {10}^{-9}\mathrm{C}\right)\left(738.0\mathrm{V}-705.0\mathrm{V}\right) \\ =8.30\times {10}^{-8\mathrm{J}} \end{gathered}$$

Therefore, the work done by the electric field on a charge that travels from B to A equals 8.30 x 10^-8 Joules.