Question

Two-point charges of equal magnitude Q are held a distance d apart. Consider only points on the line passing through both charges.

(a) 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?).

(b) Repeat part (a) for two-point charges having opposite signs.

Short answer

a) When the charges have the same signs:

(i) The potential is non-zero at any point on the line.

(ii) The electric field is zero at the mid-point of the line but non-zero at any other point.

b) When the charges have opposite signs:

(i) The potential is zero at the mid-point of the line but non-zero at any point other on the line.

(ii) The electric field is non-zero at any point on the line.

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.

When the charges are of the same nature

(a)

(i) When two charges of the same sign are held at a distance from each other, then the potential at a point having distance x=a from any charge on the line joining them is given by,

$$\begin{gathered} V_{x=a}=\frac{k{q}_1}{r}+\frac{k{q}_2}{r} \\ =\frac{KQ}{a}+\frac{KQ}{d-a}\ne 0 \end{gathered}$$

As the potential is a scalar quantity, its value at any point on the line joining centers of both the charges is equal to the algebraic sum of potentials due to individual charges, which means net potential at any point is not zero.

(ii) When two charges of the same sign are held at a distance from each other, then the electric field at the mid-point of the line joining them is equal in magnitude but opposite in direction, due to which the net electric field at that point becomes zero. In contrast, the potential at the point is still non-zero, as discussed in the previous case. Mathematically, it can be represented as:

role="math" localid="1664266032423" $$\begin{gathered} \overrightarrow{E_{mid}}=\frac{K{q}_1}{r_1^2}+\frac{K{q}_2}{r_2^2} \\ =\frac{KQ}{{\left(d/2\right)}^2}-\frac{KQ}{{\left(d/2\right)}^2}=0 \\ \overrightarrow{E_{x=a}}=\frac{k{q}_1}{r_1^2}+\frac{k{q}_2}{r_2^2} \\ =\frac{KQ}{{\left(a\right)}^2}-\frac{KQ}{{\left(d/2\right)}^2}\ne 0 \end{gathered}$$

As the electric field is a vector quantity, its value at any point on the line joining centers of both the charges is equal to the vector sum of the electric field due to individual charges, which means the net electric field at the mid-point of the line joining charges is zero while at any point is non-zero.

When the charges are of the opposite nature

(b)

(i) When two charges of opposite sign are held at a distance from each other, then the potential at mid-point and a point having distance x=a from any charge on the line joining them is given by,

$$\begin{gathered} V_{mid}=\frac{k{q}_1}{r}+\frac{k{q}_2}{r} \\ =\frac{KQ}{d/2}-\frac{KQ}{d/2}=0 \\ {V}_{x=a}=\frac{k{q}_1}{r}+\frac{k{q}_2}{r} \\ =\frac{KQ}{a}-\frac{KQ}{d-a}\ne 0 \end{gathered}$$

As the potential is a scalar quantity, its value at any point on the line joining centers of both the charges is equal to the algebraic sum of potentials due to individual charges, which means net potential at any point is zero at the mid-point and non-zero at any other point.

(ii) When two charges of opposite signs are held at a distance from each other, then the electric field at any point on the line joining both charges is non-zero compared to the potential, which is zero at the mid-point. Mathematically, it can be represented as:

$$\begin{gathered} \overrightarrow{E_{mid}}=\frac{K{q}_1}{r_1^2}+\frac{K{q}_2}{r_2^2} \\ =\frac{KQ}{{\left(d/2\right)}^2}-\frac{KQ}{{\left(d/2\right)}^2}=0 \\ \overrightarrow{E_{x=a}}=\frac{k{q}_1}{r_1^2}+\frac{k{q}_2}{r_2^2} \\ =\frac{KQ}{{\left(a\right)}^2}-\frac{KQ}{{\left(d/2\right)}^2}\ne 0 \end{gathered}$$

As the electric field is a vector quantity, its value at any point on the line joining centers of both the charges is equal to the vector sum of the electric field due to individual charges, which means the net electric field at any point on the line joining charges is non-zero.