Charge `q_(1) = +6.0n C` is on y-axis at `y = + 3cm` and charge `q_(2) = -6.0 n C` is on y-axis at `y = -3 cm`. Calculate force on a test charge `q_(0) = 2 n C` placed on X-axis at `x = 4cm`.

Options

A. `-51.8 hatj mu N`
B. `+51.8 hatj mu N`
C. `-5.18 hatj mu N`
D. `5.18 hatj mu N`

Answer

A

Explanation

Here, `q=pm6.0nC`
`=pm6.0xx10^(-9)C`
`2a=6cm=6xx10^(-2)m`
r = 4 cm (on equatorial line)
`=4xx10^(-2)m`
and `q_(0)=2nC=2xx10^(-9)C, F=?`
`F=F_(1)cos theta+F_(2)cos theta`
`=2xx(1)/(4pi epsilon_(0))(qq_(0))/(r^(2))cos theta`
`=2xx9xx10^(9)xx(6xx10^(-9)xx2xx10^(-9))/((5xx10^(-2))^(2))xx(3)/(5)`
`or F=5.18xx10^(-5)N`
Clearly, this force is along `hatj`.
`"So, "F=-51.8hatj mu N`