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# Two point charges $q_1$ and $q_2$, of magnitude $+10^{–8}\; C$ and $–10^{–8} C$, respectively, are placed $0.1\; m$ apart. Calculate the electric fields at points A, B and C shown in Figure.

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The electric field vector $E_{1A}$ at A due to the positive charge $q_1$ points towards the right and has a magnitude
$E_{1A} =\large\frac{(9 \times 10^9 Nm^2 C^{-2} ) \times (10^{-8} \;C)}{(0.05 \;m)^2}$
$\qquad= 3.6 \times 10^4 N C^{-1}$
The electric field vector $E_{2A}$ at A due to the negative charge $q_2$ points towards the right and has the same magnitude.
Hence the magnitude of the total election field $E_A$ at A is
$E_A=E_{1A} +E_{2A}$
$\quad= 7.2 \times 10^4 \;NC^{-1}$
$E_A$ is directed towards the right.
The electric field vector $E_{1B}$ at B due to the positive charge $q_1$ points towards the left and has a magnitude
$E_{1B}= \large\frac{(9 \times 10^{9} Nm^2 C^{-2}) \times (10^{-8}C)}{(0.05m )^2 }$
$\qquad= 3.6 \times 10^4 NC^{-1}$
The electric field vector $E_{2B}$ at B due to the negative charge $q_2$ points towards the right and has a magnitude
$E_{2B}= \large\frac{(9 \times 10^{9} Nm^2 C^{-2}) \times (10^{-8}C)}{(0.15m )^2 }$
the magnitude of the total electric field B is
$E_B= E_{1B} -E_{2B} =3.2 \times 10^4 \;N C^{-1}$
$E_B$ is directed towards the left.
The magnitude of each electric field vector at point C, due to charge $q_1$ and $q_2$ is
$E_{1C} =E_{2C} = \large\frac{(9 \times 10^{9} NM^2 C^{-2} ) \times (10 ^{-8} C)}{(0.10)^2}$
$\qquad = 9 \times 10^3 NC^{-1}$
The directions in which these two vectors point are indicated in the above. The resultant of these two vectors is
$E_c=E_1 \cos \large\frac{\pi}{3} + E_2 \cos \large\frac{\pi}{3}$$=9 \times 10^3 NC^{-1}$
$E_c$ points towards the right.
answered May 30, 2014 by