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Q. Two point charges $q_{A}=3 \,\mu C$ and $q_{B}=-3 \,\mu C$ are located $20 \,cm$ apart in vacuum. If a negative test charge of magnitude $1.5 \times 10^{-9} C$ is placed at this point, what is the force experienced by the test charge?

Electric Charges and Fields

Solution:

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$E_{P}=2\left[\frac{\left(9 \times 10^{9}\right)\left(3 \times 10^{-6}\right)}{0.1^{2}}\right]=5.4 \times 10^{6} N C ^{-1}$
$F_{P}=E_{P^{\prime}} q_{p}=\left(5.4 \times 10^{6}\right)\left(1.5 \times 10^{-9}\right) N =8.1 \times 10^{-3} N$