Q

# Two point charges q_A = 3 μC and q_B = –3 μCare located 20 cm apart in vacuum If a negative test charge of magnitude 1.5 × 10^–9 C is placed at this point, what is the force experienced by the test charge?

Q 1.8(b): 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?

Views

Let Q = $-1.5 \times10^{-9}$ C

The force experienced by Q when placed at O due to the charges at A and B will be:

$F = Q \times E'$

where E' is the net Electric field at the point O.

$F=1.5 \times 10^{-9} C \times 5.4 \times 10^4 N/C = 8.1\times10^{-3 } N$

Q being negatively charged will be attracted by positive charge at A and repelled by negative charge at B. Hence direction of force experienced by it will be in the direction of OA.

Exams
Articles
Questions