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
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
The electric field vector $E _{1 A}$ at $A$ due to the positive charge
$q_{1}$ points towards the right and has a magnitude $E_{1\, A }=\frac{\left(9 \times 10^{9} \,Nm ^{2} \,C ^{-2}\right) \times\left(10^{-8} \,C \right)}{(0.05 \,m )^{2}}$$=3.6 \times 10^{4}\; N\, C ^{-1}$
The electric field vector $E _{2\, A }$ at $A$ due to the negative charge $q_{2}$ points towards the right and has the same magnitude. Hence the magnitude of the total electric field $E_{ A }$ at $A$ is
$E_{ A }=E_{1 A }+E_{2 A }=7.2 \times 10^{4} \;N C ^{-1}$
$E _{ A }$ is directed toward the right.
The electric field vector $E _{1 B }$ at $B$ due to the positive charge $q_{1}$ points towards the left and has a magnitude $E_{1 B }=\frac{\left(9 \times 10^{9} \,Nm ^{2} \,C ^{-2}\right) \times\left(10^{-8}\, C \right)}{(0.05 \,m )^{2}}=3.6 \times 10^{4}\, N\, C ^{-1}$
The electric field vector $E _{2 B }$ at $B$ due to the negative charge $q_{2}$ points towards the right and has a magnitude
$E_{2 B }=\frac{\left(9 \times 10^{9} \,Nm ^{2} C ^{-2}\right) \times\left(10^{-8} \,C \right)}{(0.15 \,m )^{2}}$$=4 \times 10^{3} \,N \,C ^{-1}$
The magnitude of the total electric field at $B$ is
$E_{ B }=E_{1 B }-E_{2 B }=3.2 \times 10^{4} \,N\, C ^{-1}$
$E _{ B }$ 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_{1 c}=E_{2 c}=\frac{\left(9 \times 10^{9} \,Nm ^{2}\, C ^{-2}\right) \times\left(10^{-8}\, C \right)}{(0.10\, m )^{2}}$$=9 \times 10^{3} \,N\, C ^{-1}$
The directions in which these two vectors point are indicated in
Figure The resultant of these two vectors is
$E_{C}=E_{1} \cos \frac{\pi}{3}+E_{2} \cos \frac{\pi}{3}=9 \times 10^{3} \,N\, C ^{-1}$
$E _{ C }$ points towards the right.
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