Question

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x = 40 cm, and –60 µC at x = 60 cm, what is the magnitude of the electrostatic force on the +32-µC charge? * 48

Answer 1

Answer:

Fnet = 12 N

Explanation:

Force on a point charge due to another point charge = kq1q2 / d^2

Force on +32uC = due to + 20uC + due to -60uC

where uC = 1 x 10^-6 C and k = 9 x 10^9 N m^2 / C^2

Net Force =

$= \frac{1}{4\pi \epsilon _0} [\frac{32 \times 10^-^6 \times60\times10^-^6}{(60/100)^2}-\frac{32 \times 10^-^6 \times20\times10^-^6}{ (40/100)^2} ]$

$F_{net}=9 \times10^9\times 10^-^1^2[\frac{32\times60\times10^4}{60\times60} -\frac{32\times20\times10^4}{40\times40} ]$

$=90[32(\frac{80-60}{60\times 80} )]\\\\=90\times32\times0.004167\\\\=12N$

Fnet = 12 N