Two point charges $q_{A} = 3 μ c a n d q_{B} = - 3 μ c$ are located 20 cm apart in vaccum. What is the electric field at the midpoint of the line joining the two charges.

54 \times 10^{6} N / C a l o n g O A

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5.4 \times 10^{6} N / C a l o n g O B

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5.4 \times 10^{- 6} N / C a l o n g O B

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54 \times 10^{- 6} N / C a l o n g O B

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Solution

The correct option is B $5.4 \times 10^{6} N / C a l o n g O B$
$E = E_{A} + E_{B} = 2 E_{A}$
$= 2 [\frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}}] = 2 \times 9 \times 10^{9} \times \frac{3 \times 10^{- 6}}{10^{- 2}}$
$= 5.4 \times 10^{6} N / C a l o n s O B .$

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Two point charges qA=3μc and qB=−3μc are located 20 cm apart in vaccum. What is the electric field at the midpoint of the line joining the two charges.

The correct option is B 5.4×106N/C along OBE=EA+EB=2EA =2[14πϵ0qr2]=2×9×109×3×10−610−2 =5.4×106N/C alons OB.

Two point charges $q_{A} = 3 μ c a n d q_{B} = - 3 μ c$ are located 20 cm apart in vaccum. What is the electric field at the midpoint of the line joining the two charges.

The correct option is B $5.4 \times 10^{6} N / C a l o n g O B$
$E = E_{A} + E_{B} = 2 E_{A}$
$= 2 [\frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}}] = 2 \times 9 \times 10^{9} \times \frac{3 \times 10^{- 6}}{10^{- 2}}$
$= 5.4 \times 10^{6} N / C a l o n s O B .$