Question

An electric dipole is formed from $\pm 1.0\mathrm{nC}$charges spaced apart. The dipole is at the $2.0\mathrm{mm}$origin, oriented along the $\mathrm{x}$-axis. What is the electric field strength at the points (a) $(\mathrm{x},\mathrm{y})=(10\mathrm{cm},0\mathrm{cm})$and$\left(\mathrm{b}\right)(\mathrm{x},\mathrm{y})=(0\mathrm{cm},10\mathrm{cm})$$?$

Short answer

a.Electric field strength at point$(\mathrm{x},\mathrm{y})=(10\mathrm{cm},0\mathrm{cm})$is${\overrightarrow{\mathrm{E}}}_{\mathrm{x}}=36\mathrm{N}/\mathrm{C}\left(\mathrm{i}\right)$.

b.Electric field strength at point role="math" localid="1649114602807" $(\mathrm{x},\mathrm{y})=(0\mathrm{cm},10\mathrm{cm})$is$18\mathrm{N}/\mathrm{C}(-\mathbf{j})$.

Step-by-step solution

Formula for electric field with dipole

Moment of dipole

$\overrightarrow{\mathrm{p}}=\mathrm{q}\overrightarrow{\mathrm{d}}$

The dipole's electric field on its axis is:

${\mathrm{E}}_{\text{dipole}}\approx \frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{2\overrightarrow{\mathrm{p}}}{{\mathrm{r}}^3}$

The dipole's electric field on its bisector is:

${\mathrm{E}}_{\text{dipole}}\approx -\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{\overrightarrow{\mathrm{p}}}{{\mathrm{r}}^3}$

q is charge.

$\overrightarrow{\mathrm{d}}$is displacement vector.

$\overrightarrow{\mathrm{p}}$is dipole moment vector.

localid="1649113793496" $\mathrm{r}$is distance between the point and the dipole's centre.

Calculation of electric field (part a)

(a).

Dipole moment is,

$\overrightarrow{\mathrm{p}}=\left(1.0\times {10}^{-9}\mathrm{C}\right)\left(2.0\times {10}^{-3}\mathrm{m}\right)$

$\overrightarrow{\mathrm{p}}=2.0\times {10}^{-12}\mathrm{C}·\mathrm{m}\mathbf{i}$

At point $(\mathrm{x},\mathrm{y})=(10\mathrm{cm},0\mathrm{cm})$

Because the point is on the dipole's axis, it is on the$\mathrm{x}$-axis.

${\overrightarrow{\mathrm{E}}}_{\mathrm{x}}=\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{2\times \left(2.0\times {10}^{-12}\mathrm{C}·\mathrm{m}\right)}{(0.1\mathrm{m}{)}^3}$

${\overrightarrow{\mathrm{E}}}_{\mathrm{x}}=36\mathrm{N}/\mathrm{C}\left(\mathrm{i}\right)$

Calculation of electric field strength (part b)

(b).

At point $(\mathrm{x},\mathrm{y})=(0\mathrm{cm},10\mathrm{cm})$

Because the point is on the dipole's bisector, it is on the $\mathrm{y}$-axis.

${\overrightarrow{\mathrm{E}}}_{\mathrm{y}}=-\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{\left(2.0\times {10}^{-12}\mathrm{C}·\mathrm{m}\right)}{(0.1\mathrm{m}{)}^3}$

${\overrightarrow{\mathrm{E}}}_{\mathrm{y}}=18\mathrm{N}/\mathrm{C}(-\mathbf{j})$