91Ó°ÊÓ

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.

(a)

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

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.

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

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.

(b)

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

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.

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

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.