Question

Two point charges 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​

Answer 1

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➣Solution:

•The electric field vector $E_{1A}$ at A due to positive charge $q_{1}$ points towards the right and has a magnitude....

$E_{1A} = \frac{9 \times {10}^{9}N {m}^{2} {C}^{ - 2} \times ( {10}^{ - 8}C) }{(0.05m) {}^{2} } \\ \: \: \: \: \: \: \: = 3.6 \times {10}^{4} \: N \: {C}^{ - 1}$

•The electric field vector $E_{2A}$ at A to the negative charge $q_{2}$ points towards the right and has the same magnitude. Hence the magnitude of the total electric field $E_{A}$ A is

$E_{A} = E_{1A} + E_{2A} = 7.2 \times {10}^{4} \: N \: {C}^{ - 1}$

$E_{A}$ is directed toward 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

$E_{1B} = \frac{(9 \times {10}^{9}N \: {m}^{2} {C}^{ - 2} \times ( {10}^{ - 8}C) }{(0.05m) {}^{2} } \\ \: \: \: \: \: \: \: = 3.6 \times {10}^{4} N \: {C}^{ - 1}$

•The electric field vector $E_{2B}$ at B due to the negative charges $q_{2}$ points towards the right and has a magnitude

$E_{2B} = \frac{(9 \times {10}^{9}N \: {m}^{2} {C}^{ - 2} \times ( {10}^{ - 8}C) }{(0.15m) {}^{2} } \\ \: \: \: \: \: \: \: = 4 \times {10}^{3} N \: {C}^{ - 1}$

•The magnitude of the total electric field at 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

$E_{1C} = E_{2C} = \frac{(9 \times {10}^{9}N {m}^{2} {C}^{ - 2}) \times ( {10}^{ - 8}C) }{(0.10m) {}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 9 \times {10}^{3} N \: {C}^{ - 1}$

•The direction in which these two vectors point are indicated in the Fig. The resultant of these two vector 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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Answer 2

Answer:

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