Question 20.2: (Electric Dipole) Consider two point charges q1 = −24 nC and......
(Electric Dipole)
Consider two point charges q_{1} = −24 nC and q_{2} = +24 nC that are 10 cm apart, forming an electric dipole, see Fig. 20.6. Calculate the electric field due to the two charges at points a, b, and c.
At point a, the electric field vector due to the negative charge q_{1}, is directed toward the left, and its magnitude is:
E_{1a}=k\frac{|q_{1}|}{r_{1a}^{2}}=(9\times10^{9}\,\mathrm{N{\mathrm{.}}}\mathrm{m}^{2}/C^{2})\frac{(24\times10^{-9}\,\mathrm{C})}{(0.04\,\mathrm{m})^{2}}=135\times10^{3}\,\mathrm{N{\mathrm{/C}}}The electric field vector due to the positive charge q_{2} is also directed toward the left, and its magnitude is:
E_{2a}=k{\frac{|q_{2}|}{r_{2a}^{2}}}=(9\times10^{9}\,\mathrm{N{.m^{2}}/C^{2}}){\frac{(24\times10^{-9}\,\mathrm{C})}{(0.06\,\mathrm{m})^{2}}}=60\times10^{3}\,\mathrm{N/C}Then, the resultant electric field at point a is toward the left and its magnitude is:
E_{a}=E_{1a}+E_{2a}=135\times10^{3}\,\mathrm{N}/\mathrm{C}+60\times10^{3}\,\mathrm{N}/\mathrm{C}= 195 × 10³ N/C (Toward the left)
At point b, the electric field vector due to the negative charge q_{1}, is directed toward the left, and its magnitude is:
E_{1b}=k\frac{|q_{1}|}{r_{1b}^{2}}=(9\times10^{9}\,\mathrm{N{\mathrm{.}}}m^{2}/C^{2})\frac{(24\times10^{-9}\,\mathrm{C})}{(0.12\,\mathrm{m})^{2}}=15\times10^{3}\,\mathrm{N{\mathrm{/C}}}In addition, the electric field vector due to the positive charge q_{2} is directed toward the right, and its magnitude is:
E_{2b}=k{\frac{|q_{2}|}{r_{2b}^{2}}}=(9\times10^{9}\,\mathrm{N{.m^{2}/C^{2}}}){\frac{(24\times10^{-9}\,{\mathrm{C}})}{(0.02\,{\mathrm{m}})^{2}}}=540\times10^{3}\,\mathrm{N/C}Since E_{2b} > E_{1b}, the resultant electric field at point b is toward the right and its magnitude is:
E_{b}=E_{2b}-E_{1b}=540\times10^{3}\,\,\mathrm{N}/\mathrm{C}-15\times10^{3}\,\,\mathrm{N}/\mathrm{C}= 525 × 10³ N/C (Toward the right)
At point c, the magnitudes of the electric field vectors \vec{E}_{1c} and \vec{E}_{2c} established by q_{1} and q_{2} are the same because |q_{1}| =| q_{2}| = 24 nC and r_{1c} = r_{2c} = 10 cm.
Thus:
The triangle formed from q_{1}, q_{2}, and point c in Fig. 20.6 is an equilateral triangle of angle 60°. Hence, from geometry, the vertical components of the two vectors \vec{E}_{1c} and \vec{E}_{2c} cancel each other. The horizontal components are both directed toward the left and add up to give the resultant electric field E_{c} at point c, see the figure below.
Thus: E_{c} = E_{1c} cos 60° + E_{2c} cos 60° = 2E_{1c} cos 60°
= 2(21.6 × 10³ N/C)(0.5) = 21.6 × 10³ N/C (Toward the left)
